TG 


JC-NRLF 


THEORY 


lAROHBS. 


Prof.    W.    ALLAN, 


Formerly  of  Washington  <nifi  I.w  University,  ' 


NEW   YORK: 

D.    VAN    NOSTRAND    COMPAN.i', 
•^3  MURRAY  STREET  AND  27  WARREN  STREET. 

1890, 


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THEORY 


ARCHES 


BY 

Prof.    W.    ALLAN, 

Formerly  of  Washington  and  Lee  University,  Lexington,  Fa. 


NEW   YORK: 
D.    VAN    NOSTRAND    COMPANY, 

23  MURRAY  STREET  AND  27  WARREN  STREET. 

1890. 


COPYRIGHT. 
D.  VAN  NOSTRAND  COMPANY. 

1890. 


PREFACE  TO  SECOND  EDITION. 


The  original  edition  of  this  Mono- 
graph was  reprinted  from  the  pages  of 
"  Van  Nostrand's  Engineering  Maga- 
zine," to  which  it  was  contributed  by  the 
late  Prof.  W.  Allan.  It  had  been  pri- 
marily prepared  by  him  from  a  series  of 
notes,  which  notes  had  been  litho- 
graphed for  the  use  of  his  classes  in 
studying  "Kankine's  "  works. 

In  printing  this  second  edition,  there- 
fore, it  has  not  been  thought  necessary 
to  make  any  changes  whatever,  as  the 
text  is  simply  an  amplification  and  ex- 
planation of  the  "  Theory  of  Arches  "  as 
given  by  Prof.  Eankine,  and  as  Prof. 
Eankine  himself  has  been  dead  some 
years,  his  treatment  of  the  subject  as  de- 
veloped in  his  Manual  is  probably  as 

127692 


complete  as  it  ever  will  be.  It  would 
seem,  therefore,  that  this  little  book  will 
answer  the  purposes  for  which  it  was 
intended  for  all  time  to  come.  To  the 
reader  and  student  of  Kankine's  works  it 
will  undoubtedly  be  found  interesting  as 

well  as  useful. 

THE  PUBLISHERS. 
May,  1890. 


THEORY  OF  ARCHES. 


The  following  is  an  amplification  and 
explanation  of  Professor  Rankine's  chap- 
ters on  this  subject. 

Perhaps  the  clearest  way  of  developing 
the  "  Theory  of  Arches  "  is  to  begin  with 
the  consideration  of  the  forces  which  act 
upon  a  suspended  chain  or  cord.  The 
force  in  the  chain  or  cord  is  just  the  op- 
posite of  that  upon  an  arch—  that  is,  it 
is  tension  instead  of  compression,  but 
the  relations  between  the  "external" 
and  "  internal "  forces,  or,  what  is  the 
same,  between  the  loads  and  the  resist- 
ances they  produce,  are  strictly  analo- 
gous. 

Let  CAB  (Fig.  1)  be  a  cord  suspend- 
ed at  C  and  B  and  loaded  in  any  manner 
over  its  whole  length.  Consider  the 
forces  acting  on  this  cord.  Suppose  it 


6 


attached  to  a  hook  at  B  and  to  another 
at  C.  A  cord  without  stiffness  cannot 
exert  a  pull  except  in  the  direction  of  its 
length;  therefore  the  "pulls"  in  the 
rope  at  C  and  B,  and  exerted  at  these 
points  on  the  suspending  hooks,  must 
be  in  the  direction  of  the  tangents  at 
FIG.  1. 


those  points.  The  load  is  supposed  to  be 
distributed  over  the  cord,  but  we  may 
find  its  resultant.  Let  P  be  this  result- 
ant and  P  F  its  direction.  The  three 


forces,  viz.,  the  pulls  at  C  and  B,  and 
the  resultant  of  the  load,  P,  are  all  in 
the  same  vertical  plane ;  they  are  the 
only  forces  acting  on  the  cord ;  and  as 
they  are  in  equilibrium,  the  directions  of 
these  three  forces  must  meet  in  one  point, 
and  the  forces  themselves  must  be  pro- 
portional to  the  three  sides  of  a  triangle 
drawn  parallel  to  their  directions. 

G  N  F  (Fig.  1)  is  such  a  triangle.  The 
known  directions  of  the  pulls  at  B  and 
C,  and  of  P,  give  us  the  angles  in  this 
triangle  ;  and  if  we  know  also  the  mag- 
nitude of  the  load  P,  represented  by  the 
line  G  F,  we  can  determine  that  of  the 
pulls  at  B  and  C.  For 

(PullatB=GN)  :GF  :   rsinGFN  isinGNF. 

(PullatC=NF)  :  GF  :    isinNGF  :  sinGNF. 

The  analysis  we  have  made  for  the 
whole  cord  may  be  applied  to  any  part 
of  it.  Thus,  if  we  consider  any  arc  B' 
A'  (Fig.  2)  of  the  cord,  and  the  load  on 
that  arc,  we  have  three  forces  in  the 
same  plane  in  equilibrium.  For  at  A' 
and  B'  the  other  parts  of  the  cord  may 


8 

be  replaced  by  two  hooks,  and  the  pulls 
on  these  hooks,  exerted  by  the  cord  at  A' 
and  B',  will  be,  as  before,  in  the  direction 
of  the  tangents  at  those  points.  The  re- 
sultant P'  of  the  load  on  A'  B'  must 
pass  through  the  point  of  intersection 
of  the  tangents,  and  if  the  direction  of 
that  resultant  be  as  indicated  in  the  fig- 
ure, then  G'  N'  F'  will  be  the  triangle  of 
forces. 

FIG.  2. 


The  principles  above  explained  enable 
us  to  calculate  the  "  pulls  "  at  all  points 
of  a  loaded  chain  or  cord,  and  conse- 
quently to  fix  its  size  and  strength  to 
bear  a  given  load  ;  or  to  determine  the 
amount,  distribution  and  direction  of  the 
load  necessary  to  produce  assumed 
"  pulls  "  in  a  cord  of  a  given  shape. 


Thus,  suppose  in  the  half  of  the  loaded 
cord  of  (Fig.  1)  we  draw  the  tangents  at 
A  and  B  (as  is  done  in  Fig.  3),  the  re- 
sultant of  the  load  P  must  pass  through 
F,  the  point  of  intersection  of  the  tan- 
gents. If  the  direction  and  amount  of 
P  be  known,  lay  off  F  G  to  represent  it. 

FIG.  3. 


X 


Then,  as  above  explained, 

N  F  =  pull  at  A 
and 

N  G  =  pull  at  B. 

Suppose,  on  the  other  hand,  we  as- 
sume the  pulls  at  A  and  B  to  be  equal, 
we  lay  off  on  the  two  tangents  (Fig.  4) 
equal  lengths,  F  S  and  F  N,  to  repre- 
sent these  equal  pulls,  and  upon  them 
construct  a  parallelogram.  Then  F  G 


10 


gives  the  magnitude  and  direction  of  the 
resultant  of  the  load  that  must  be  put 
on  the  cord  to  produce  the  given  pulls. 
A  cord  is  in  equilibrium  when  it  is 
balanced  under  the  load  applied.  Change 
the  distribution  of  the  load  and  the 

FIG.  4. 


cord  at  once  changes  shape  and  as- 
sumes the  form  necessary  to  equilib- 
rium under  the  new  load. 

Thus,  if  P  (Fig.  5)  equals  the  direc- 
tion of  the  resultant  of  the  new  load  on 
the  cord  from  the  horizontal  point  A  to 
the  point  of  support  B,  draw  the  tan- 
gent A  F,  until  it  meets  the  direction  of 
the  load  P,  at  F  ;  then  draw  F  B.  The 


11 


cord  A  B  will  have  so  changed  its  form 
that  F  B  (Fig.  5)  will  now  be  the  direc- 
tion of  the  tangent  at  B. 

FORMS    OF    CORDS    UNDER    VARIOUS    LOADS. 

Let  us  now  investigate  the  various 
curves  which  a  cord  will  assume  under 
different  distributions  of  the  load. 

FIG.  5. 


Case  I.  Suppose  the  load  be  alto- 
gether vertical,  and  to  be  distributed 
uniformly  along  the  horizontal. 

Let  equal  weights  be  hung,  for  in- 
stance, along  a  cord  C  B  (Fig.  6)  so  that 
the  horizontal  distance  between  the 


12 

threads  by  which  the  weights  are  sus- 
pended shall  be  everywhere  equal.  Or, 
draw  little  elementary  triangles  along 
the  curve,  so  that  the  bases  of  all  these 
little  triangles  shall  be  equal,  and  let  the 
threads  holding  the  weights  cut  the  mid- 
dle of  these  bases.  Then  each  weight 
FIG.  6. 


may  be  considered  as  the  resultant  of 
the  load  on  'the  element  of  the  curve 
which  constitutes  the  hypothenuse  of 
the  little  triangle  to  which  it  is  attached. 
Such  a  load  is  vertical  and  is  uniformly 
distributed  along  the  horizontal. 

To  determine  the  curve  of    the  cord. 
Obtain  the  resultant  of  the  load  between 


13 

the  horizontal  point  A  and  the  point  B 
(Fig.  7).  This  resultant,  as  the  little 
forces  are  all  parallel,  is  equal  to  the  sum 
of  them,  and  it  is  vertical  in  direction. 

Fia.  7. 


It  will  also  evidently  bisect  A  T.  Draw 
it,  and  from  its  point  of  intersection 
with  A  T  draw  the  line  F  B,  which,  as 
has  been  shown,  must  be  tangent  at  B. 
Prolong  B  F  to  I,  then  the  subtangent 
I  J  is  seen  to  be  bisected  at  the  vertex 
A  of  the  curve.  Hence  the  curve  C  B 
is  a  parabola. 


14 


The  triangle  B  F  T  has  its  sides  par- 
allel to  the  forces  acting  on  the  half  cord 
A  B  ;  so  that  if  B  T  be  taken  to  repre- 

sent P, 

B  F  =  pull  at  B 
F  T  =  pull  at  A. 
Let  T  =  equal  tension  at  any  point  along  the 

cord. 

H  =  value  of  T  at  the  horizontal  point 
A,  or  the  "  horizontal  pull"  on  the 
cord. 

i  =  inclination  of    the  tangent  at   any 
point  to  the  horizontal. 

Then  as  the  arc  A  F  (Fig.  7)  may 
stand  for  any  part  of  the  curve  counting 
from  the  horizontal  point  A  towards  one 
of  the  points  of  suspension,  we  have  the 
following  general  equations  from  the 
triangle  B  F  T  : 

(1.) 


Tan  i--.-  *JL  (2) 

-  H  -   H  -  dx 

(p  being  =  the  load  per  unit  of  horizon- 
tal distance,  A  the  origin  of  co-ordinates, 
A  T  =  axis  of  X  and  A  J  =  axis  of  y). 

From  equations  (1)  and  (2)  we  can 
solve  three  problems. 

1.  Given  the  curve,  and  the  load,  to 
find  T  and  H. 


15 

2.  Given   the  curve,  and  T  and  H,  to 
find  P. 

3.  Given  the  load,   and  T  and  H,  to 
find  the  curve. 

For  a  full  discussion  of   this  case,  see 
Rankine's  "Civil  Engineering." 

Such  a   distribution  of  the  load  as  we 

FIG.  8. 


have  discussed  in  the  above  case  is  ap- 
proximated to  in  suspension  bridges,  and 
sometimes  in  wood,  iron,  or  steel  arches, 
but  not  usually  in  stone  or  brick  ones. 


16 

Case  II.  Let  the  load  still  be  vertical, 
bnt  distributed  uniformly  along  the 
curve. 

That  is,  divide  the  arc  C  A  B  (Fig.  8) 
into  elements  each  of  a  unit  in  length  ; 
then  the  load  on  these  elements  is  constant 
throughout.  It  is  easily  seen  that  such  a 
load  is  not, as  in  the  last  case,uniform  along 
the  horizontal,  for  the  bases  of  the  little 
triangles  of  which  the  hypothenuses  are 
now  equal,  diminish  in  extent  as  we  go 
from  A  towards  B  or  C.  A  chain  of  uni- 
form material  and  cross-section,  and  act- 
ed on  by  nothing  but  its  own  weight,  is 
in  the  condition  described,  and,  as  is 
well  known,  the  curve  assumed  by  it  is 
the  "  common  catenary." 

Let  p  =  weight  of  a  unit's  length  of 
the  cord,  then  if  p  m  =  horizontal  pull 
on  the  cord  at  A  =  H,  m  is  called  the 
modulus  of  the  catenary,  and  represents 
the  length  of  cord  of  the  same  kind  as 
C  B,  the  weight  of  which  would  equal 
the  pull  at  A.  The  weight  on  A  B  =  P 
=  p  s  when  s  —  length  of  cord  A  B. 

The  triangle  of  forces  for  any  arc  A  D 


17 

(Fig.  9)  can  be  found  as  before,  by  drawing 
the  tangents  at  A  and  D,  and  the  line  rep- 
resenting the  force  P  vertically  through 
their  intersections.  The  triangle  D  F  T 
will  represent  the  forces  ;  D  T  being  =  P 
=  p  s,  and  F  T  =  H  =  p  m,  and  D  F  = 
T  =  tension  at  D.  Then 
FIG.  9. 


F  T     pm      m     d  x 

From  the  differential  equation 
d  y_  s 
d  x      m 


(4.) 


18 

we  obtain  the  linear  equation  of  the 
curve.  In  doing  so  it  is  most  conveni- 
ent to  take  the  origin  at  a  point  O, 
whose  distance  below  the  vertex  A  is  = 
m.  The  line  Q  O  X  (Fig.  10)  is  called 
the  directrix  of  the  catenary. 

The  equations  of  the  catenary  are 

IMj      )          X  'X-     )  i/7/2_|_j»j2 

:~2~(E™  -E       m-']  •~iengthofaro»(5.) 

_  m   (    _%_ x_ ) 


x  =  m.  hy.  log.  j  j,  /  #»  V.         (7.) 

—  —  )—  /i/  .  —  —  -  —  j 
w       r     m9 

Area  AGE  D=fyxd  —  m  s  (8.) 


r>    j-  -- 

Radius  of  curv.  =  p  =—  -  —  =  —  ,.  A  . 

m8  m  (10.) 

Since  the  area  A  O  E  D  —  m  s,  and 
m  =  a  constant,  the  area  varies  as  s. 
But  the  load  on  the  arc  A  D  (=p  s)  also 

*JS  =  Base  of  Naperian  Logarithms. 


19 


varies  as  s,  since  p  is  constant.  Hence  a 
convenient  mode  of  representing  the  load 
on  any  arc,  A  D^  Suppose  a  sheet  of 


metal  C  Q  T  B  A  C  (Fig.  10),  bounded 
below  by  the  "directrix,"  Q  T,  to  be 
suspended  from  the  curve.  Let  the 
weight  of  this  metal  corresponding  to  m 


20 

units  of  its  surface  be  =  p.  That  is,  let 

P  . 
w  m  —  p,  or  w  —    m 

The  weight  of  a  strip  a  unit  in 
breadth  extending  from  A  to  O  is  then 
=  p  =  the  weight  of  a  unit's  length  of 
the  cord.  Then  the  part  of  the  sheet  A 
ODE  whose  weight  —  w  m  s  =  p  s  rep- 
resents the  weight  P  on  the  arc  A  D.  So 
A  O  B  T  represents  the  weight  on  A  B, 
and  C  Q  T  B  the  whole  weight  on  C  A 
B.  In  the  horizontal  pull  at  A  we  have 

H  =  p  m  —  w  m2  (!!•) 

and  at  any  pointD 


T=  i/H2-]-P2  =p  \/s*-\-m*=py=w  my.  (12.) 

The  property  above  explained  may  be 
illustrated  in  another  way. 

Construct  on  A  B  (Fig.  11)  a  series  of  lit- 
tle triangles  with  all  their  bases  equal.  Let 
the  weights  of  the  little  arcs  constituting 
the  hypothenuses  of  these  triangles  be 
represented  by  balls  suspended  by 
threads  from  the  middle  of  each  little 
arc.  Take  the  length  of  the  thread  cor- 
responding to  the  ball  at  A  as  =  m  ; 


21 

make  the  lengths  of  all  the  threads  pro- 
portional to  the  weights  of  the  balls  hung 
to  them ;  then  the  lower  ends  of  these 
lines  will  all  be  on  the  directrix  O  X. 

FIG.  11. 


That  is,  the  intensity  of  the  load  on  a 
catenary  along  the  horizontal  line  (  — 
weight  on  a  unit  of  horizontal  distance) 
varies  as  the  ordinates  of  the  catenary, 
when  those  ordinates  are  measured 
from  the  directrix. 


22 


It  makes  no  difference  in  the  form  of 
the  curve  A  B  (Fig.  11),  to  increase  or 
diminish  the  weights,  provided  the  pro- 
portion among  them  is  preserved.  Thus 
we  may  assume  the  cord  and  the  sheet 
C  Q  T  B  (Fig.  10),  to  be  of  a  different 
material  in  which  a  unit's  length  of  the 
cord  shall  in  weight  =p',  and  the  weight 
of  the  sheet  per  unit  of  surface  shall  = 
w',  and  A  B  will  be  unchanged.  Note, 
however,  that  we  cannot  change  the 
depth  A  O  of  the  sheet  (Fig.  10),  nor  the 
length  of  the  lines  (Fig.  11),  without 
changing  the  curve ;  for  if  the  lines  ended 
in  O'  X'  for  instance,  instead  of  O  X,  then 

A  O  would  not  be  equal  to  A  O'. 
D  E  D  E' 

Hence,  the  modulus  (m  =  A  O)  fixes 
the  catenary,  or,  if  we  assume  the  catena- 
ry, this  determines  the  modulus.  Thus, 
if  we  assume  three  points,  B,  A,  C  (Fig. 
10),  on  the  catenary,  the  distance  A  O  is 
thereby  determined  ;  and  if  we  assume 
A  O  and  the  point  A  we  cannot  generally 
assume  B  and  C. 

This  often  interferes  with  the   use  of 


23 

the  "  common  catenary  "  in  the  building 
of  arches  [in  which  case  the  curve  is  in- 
verted, the  metal  sheet  A  O  T  D  is  re- 
placed by  a  wall  of  uniform  material,  and 
the  tension  on  its  cord,  C  B  (Fig.  10),  is 
replaced  by  a  thrust  along  CAB  (Fig. 

FIG.  12. 

r 


12)].  For  we  are  often  compelled  to 
make  the  curve  pass  through  three 
points,  while  yet  the  value  of  A  O  is 
fixed. 

But  this  difficulty  may  be  obviated  by 
the  use  of  the  transformed  catenary, 
which  we  will  now  discuss. 


24 

'Case  III.  By  the  principle  of  Parallel 
Projections,  if  any  cord  or  arched  rib  is 
'balanced  under  a  system  of  forces  which 
=are  represented  in  the  figure  by  lines,  and 
=a  parallel  projection  be  made  of  the 
<;urve  of  the  cord  or  rib  and  of  the  lines 
representing  the  forces,  then  the  new 
curve  will  represent  a  cord  or  rib  that 
will  be  balanced  under  the  forces  repre- 
sented by  the  new  lines. 

Imagine  a  cylindrical  surface  con- 
structed upon  CQ  T  B  A  C  (Fig.  10)  as  a 
base.  To  simplify  matters,  suppose  the 
elements  of  the  cylinder  to  be  perpen- 
dicular to  the  plane  of  the  base.  Cut 
this  cylinder  by  a  plane  inclined  to  the 
base,  and  we  shall  get  a  "  Transformed 
Catenary,"  and  the  shape  of  the  sheet 
of  metal  under  which  it  will  be  balanced  ; 
for  the  new  curve  and  surface  cut  out 
by  the  inclined  plane  are  the  parallel 
projections  of  the  curve  CAB  and  the 
surface  C  Q  T  B  A  C  (Fig.  10).  Let  this 
inclined  plane  be  so  placed  that  it  shall 
intersect  the  plane  of  the  base  in  the 
straight  line  C  B  (Fig.  10)  or  in  one  paral- 


25 

lei  to  it.  Then  all  horizontal  lines  (or 
those  parallel  to  C  B  or  Q  T)  will  be  un- 
changed in  length  in  the  parallel  projec- 
tion, while  all  vertical  lines  (those  parallel 
to  AO,  etc.,)  will  be  lengthened  in  a  con- 
stant ratio  whose  magnitude  will  depend 

FIG.  13  (a). 


upon  the  inclination  of  the  cutting  plane. 
Make  a  vertical  section  of  the  cylinder 
on  the  line  O  Y.  Then  if  the  cutting 
plane  passes  through  C  B  we  get  the 
triangle  O  U  Y  (Fig.  13a)  cut  out  of 
the  wedge  to  which  the  cylinder  reduces 


26 

in  this  case.     In  the  triangle,  U  Y  is  the 
ordinateof  the  vertex  of  the  transformed 

FIG.  13  (b). 


catenary,  corresponding  to  O  A  in  the 
common  catenary,  and  all  lines  parallel 
to  U  V  are  evidently  increased  over  the 


27 

corresponding  ones  of  which  they  are  the 
parallel  projections,  in  the  same  ratio  that 
U  V  exceeds  O  A.  Laid  down  in  the 
same  plane  the  two  curves  are  CAB 
and  C'  A'  B'  (Fig.  13  b). 

It  is  easy  to  pass  from  a  given  cate- 
nary to  a  transformed  catenary  whose 
ordinates  shall  be  shorter  instead  of 
longer  than  those  of  the  given  curve,  by 
erecting  an  oblique  cylinder  on  the  given 
catenary  and  surface  C  Q  T  B,  and  cut- 
ting it  by  a  plane  less  oblique  than  the 
base.  So,  too,  the  horizontal  dimen- 
sions can  be  changed  instead  of  the  ver- 
tical, by  making  the  cutting  plane  meet 
the  base  in  a  line  parallel  to  O  Y,  instead 
of  in  one  parallel  to  Q  T. 

The  equations  of  the  curve  C'  A'  B' 
(Fig.  136)  are  ttfus  obtained.  The  ab- 
scissas are  the  same  as  those  in  C  A  B, 
but  the  ordinates  are  changed,  so  that  (if 
y  =  general  ordinate  of  C'  A'  B'  and  y0 
—  A'  O,  the  ordinate  at  the  vertex  A') 

y'  :  y  :  :  A'  o  :  A  o  :  :  y0  :  m. 

•'•*'=  IT  •yny-y'-^ 


28 

In  the  equations  of  the  common  cat- 
enary substitute  y'  for  y  and  we  have 
the  equations  of  C'  A'  B'. 

From  equation  (6) 

ra      ,_  ra  (     j£_  j*L  I 

-fc  *=-*-'•  (V3  +E    ~  m  ) 

•      y'    _   I!*!        A.  -    ^       (. 

"    2    (E  m  +  E       w  f  (13.) 

So  equation  (7)  becomes 


*  =  «thy.  log. 


So  equation  (8)  or  area  A'  O  E  D'. 

(15.) 

etc.,  etc.,  etc. 

The  "  triangle  of  forces  "FED  (Fig. 
135),  for  any  arc  A  D  of  the  catenary,  be- 
comes FED'  for  the  arc  A'  D'  of  the 
transformed  catenary — that  is,  since  the 
horizontal  lines  and  forces  are  un- 
changed. 

Tension  at  vertex  A'  =  H'  =  H  =wm*      (16.) 


29 


Load  on  A'  D'  is  increased  in  ratio  of 
A'  O  to  A  O  or  of  D'  E  to  D  E. 


A'D  E~    '    m  (17.) 

(D'  E  represents  this  load.) 
Then  tension  at  D'  is 

T'  =  |/p»  H_  H*  (18«) 

and 

~&x  ~  2m  I E  #i    —  E~  ~wj  (19.) 

In  this  curve  we  can  assume  the  direc- 
trix Q  T,  the  distance  A'  O  (  =  y0)  and 
also  the  points  B'  and  C'.  These  quan- 
tities assumed,  we  determine  m  (the 
modulus  of  the  corresponding  common 
catenary)  from  equation  (14)  and  then  by 
equation  (13)  find  points  of  the  trans- 
formed catenary. 

From  equations  (18)  and  (19)  we  can 
solve  three  problems  similar  to  those 
given  under  the  head  of  Case  I. 

Case  IV.  So  far  we  have  discussed 
the  forms  of  cords  under  loads  par- 
allel and  altogether  vertical.  Let  us 


30 

take  up   the  cases  of  loads   varying  in 
direction. 

Suppose  (as  Case  IV.)  that  the  load  be 
uniform  and  normal  at  every  point  to 
the  cord.  Such  a  load  is  represented  in 
(Fig.  14),  the  load  on  each  element  d  s 
of  the  curve  being  constant  and  perpen- 
dicular to  it. 

FIG.  14. 


It  is  first  to  be  noted  that  the  pull  or 
tension  on  a  cord  under  any  load  which 
is  everywhere  normal  to  it,  must  be  con- 
stant. That  is,  the  pull  along  the  cord 
at  A  and  B,  and  at  all  other  points,  is 
one  and  the  same.  That  the  tension  at 


31 

B  in  the  cases  previously  discussed  is 
greater  than  at  A,  is  due  to  the  fact  that 
the  elements  of  the  load  between  A  and 
B  have  in  those  cases  tangential  com- 
ponents, which  go  to  change  the  value  of 

FIG.  15. 


the  pull  along  the  cord.  But  in  the 
present  case,  the  load  being  everywhere 
normal,  there  are  no  such  tangential  com- 
ponents, and  therefore  the  "  pull  "  does 
not  change. 


32 

Take  any  two  adjoining  elements  of 
the  cord  d  s  (=  D  E,  Fig.  15)  and  d  s> 
(=  E  F,  Fig.  15),  each  of  such  length 
as  to  correspond  to  equal  elements  of  the 
load.  The  little  loads  on  these  lines  we 
will  represent  by  p  d  s,  and  p'  d  s'.  Note 
that,  unless  the  load  be  uniform  all 
around  the  cord,  d  s  will  not  be  equal 
to  d  s'.  The  equal  loads  p  d  s  and  p'  d 
s'  being  normal  respectively  to  D  E  and 
E  F,  their  resultant  which  lies  in  the 
direction  O  K  (Fig.  15)  bisects  the  angle 
between  p  d  s  and  p  d  s',  and  also  the 
angle  D  E  F  between  d  s  and  d  s',  which 
last  is  the  angle  between  the  direction 
of  the  pulls  T  and  T'  on  the  cord  at  D 
F.  Hence  the  parallelogram  of  forces 
(  as  shown  at  R")  will  be  a  rhombus,  or 


Again,  take  three  elements,  D  E,  E  F, 
F  H  (Fig.  16),  of  the  cord,  each  bearing 
the  normal  load  pds=pds  =  pd  s. 
In  place  of  the  little  arcs,  we  use  for 
clearness  the  chords  of  those  arcs. 
Since  the  load  around  the  whole  curve 


33 

CAB  (Fig.  14)  is  supposed  to  be  uni- 
form, the  arcs  bearing  the  equal  ele- 
ments (p  d  s)  of  that  load  must  also  be 
equal,  or  DE  =  E  F  =  F  H.  We  have 
above  proved  T  —  T".  Hence  the  three 
sides  D  E,  E  F,  and  F  H  will  arrange 
themselves  symmetrically  as  in  (Fig.  16). 

FIG.  16. 


Now,  every  other  piece  of  the  cord  con- 
taining three  elements  will  assume  ex- 
actly the  same  slope  as  D  H,  since  each 
such  piece  must  equal  D  H  in  length 
and  must  be  acted  on  by  an  equal  and 
precisely  similar  system  of  forces.  Con- 
sequently, the  little  chords  D  E,  etc.r 
must  constitute  a  regular  polygon,  and 


34 

the  curve  in  which  they  are  inscribed 
must  be  constant  in  curvature,  in  other 
words — a  circle. 

Therefore  the  curve  of  the  cord  CAB 
(Fig.  14)  is  the  arc  of  a  circle. 

To  form  the  triangle  of  forces  for  any 
point  of  a  loaded  circle  as  for  A  D  (Fig. 

FIG.  17. 


X 


x  P 

17),  draw  the  tangents  at  the  extremities 
A  and  D.  From  the  intersection,  F,  of 
these,  lay  off  F  N  =  F  S,  to  represent 
the  equal  pulls  at  A  and  D.  Then  the 
diagonal  F  G  =  the  resultant  of  the 
load,  and  the  triangle  F  N  G  or  F  S  G 
represents  the  forces  acting  on  A  D. 

It  is  often  easier  to  deal  with  a  uni- 
form normal  load  by  resolving  it  into 
its  vertical  and  horizontal  components. 


35 

The  load  on  an  element  D  ~Et=ds  of  the 
quadrant  A  B  (Fig.  18)  is  =  p  d  s.  The 
horizontal  component  of  this  load  = 
p  d  s  sin.  6^  where  6  =  the  angle  made 
by  the  direction  of  p  d  s  with  the  verti- 

FIG.  18. 


cal  (or  what  is  the  same,  the  angle  made 
by  the  tangent  of  ds  with  the  horizontal). 
The  vertical  component  =  p  d  s  cos.  6. 
Consider  the  horizontal  component  (p  ds 
sin.  6)  with  reference  to  the  vertical 
space  over  which  it  is  distributed.  This 


36 

space  is  E  K  (Fig.  18)  =  d  s  sin.  6. 
Hence  the  intensity  of  the  horizontal 
component 

_p  d  s  sin.  e 

~~  d  s  sin.  e 

So  the  vertical  component  (p  d  s  cos.  6)  is 
distributed  over  a  horizontal  space  =  D  K 
—  d  s  cos.  6,  and  therefore  its  intensity 
is 

pdscos.o 


d  scos.o 


=P> 


But  p  =  the  intensity  of  the  normal 
force.  Hence  the  original  normal  force 
at  each  point  is  equivalent  to  a  horizon- 
tal and  a  vertical  force,  at  that  point,  of 
equal  intensity. 

If  we  then  construct  little  triangles  on 
the  curve  A  B  (Fig.  19)  such  that  their 
vertical  sides  shall  be  constant  in  length, 
the  horizontal  forces  on  these  sides  will 
be  represented  by  lines  of  constant 
length.  Transfer  these  forces  in  their 
lines  of  direction  to  A  Y.  A  Y  is  the 
sum  of  all  the  vertical  sides  of  the  lit- 
tle triangles,  and  as  the  horizontal  inten- 
sity is  constant  and  equal  to  p,  we  have 


37 

(if  r  —  radius  of  the  circle)  p  (A  Y)  ~  p  r 
=  total  horizontal  force  on  quadrant 
AB. 

Similarly,  if  we  draw  a  set  of  triangles 
on  A  B  with  all  their  horizontal  sides  of 


the  same  length,  we   may  see   that   the 
total  vertical  force  on  A  B  is 


Hence, 

1.  The  resultant  of  the  entire  normal 


38 


force  on  the  quadrant  A  B  is  equal  to  the 
resultant  of  a  horizontal  and  a  vertical 
force  each  of  which  is  —  p  r. 

2.  Therefore  in  the   parallelogram  of 

forces  for  the  quadrant    (Fig.  20),  F  S, 

which  represents  the  pull  along  the  cord 

at   B,   is   the  vertical   component  of  P, 

FIG.  20. 


while  N  F  =  pull  at  A,  is  the  horizontal 
component  of  P.  Each  of  these  forces  = 
p  r.  Therefore  the  constant  pull  all 
along  the  cord  is  ==  p  r. 

If   we   make  the  pull   at   the  vertical 
point  (B)  =  V,  we  have 

H  =  V  =  T  =  pr        .        .        (20.) 


39 

In  practice  a  uniform  normal  force  ex- 
ists in  the  case  of  a  cylinder  filled  with 
steam,  or  in  a  vertical  cylinder  filled 
with  liquid.  Thrust  instead  of  tension 
along  A  B  exists  when  the  normal  force 
pushes  inwards,  as  in  the  tubes  of  a 
steam  boiler  or  an  empty  vertical  cylin- 
der immersed  in  water.  In  reference  to 
arches,  this  discussion  has  its  princi- 
pal value  as  introductory  to  those  that 
follow. 

Case  V.  In  this  case  we  obtain  the 
curve  and  forces  by  parallel  projections 
from  the  circle. 

If  we  suppose  a  cylinder  erected  upon 
the  circle  (Fig.  21)  as  abase,  and  cut  it  by 
an  inclined  plane  whose  line  of  intersec- 
tion with  the  plane  of  the  base  shall  be 
parallel  to  A  I,  we  will  get  an  ellipse 
whose  vertical  axis  A'  I'  (Fig.  21)  will  = 
A  I,  and  whose  horizontal  axis  C'  Br  will 
be  greater  than  C  B.  All  lines  parallel  to 
A  I  will  be  unchanged  in  length,  while 
all  parallel  to  C  B  will  be  increased  in 
the  proportion  of  C'  B'  to  C  B.  Now, 
by  the  principle  of  parallel  projections, 


40 

the  ellipse,  which  is  the  parallel  projec- 
tion of  the  circle,  will  be  balanced  under 
the  forces  which  are  the  parallel  projec- 
tions of  those  under  which  the  circle  is 
balanced. 

As  we  have  seen,  the  circle  is  the  curve 
assumed  by  the  ring  under  a  uniform 
horizontal  and  vertical  force  at  each 
point  of  the  same  kind,  and  equal  in  in- 
tensity; for  such  a  system  of  forces  is 
equivalent  to  a  constant  normal  force 
around  the  curve.  For  convenience, 
these  forces  are  represented  in  Fig.  (21) 
along  the  two  diameters,  each  little  line 
representing  the  force  on  a  unit  of  dis- 
tance. The  pull  around  the  ring  is  of 
course  tangential  to  it,  and  is  everywhere 
the  same  (  =  p  r).  This  pull  is  repre- 
sented at  A  and  B  by  the  arrows 
there. 

In  the  ellipse,  the  vertical  lines  being 
unchanged,  the  total  vertical  force  on  the 
elliptic  ring  (=  the  sum  of  all  the  little 
vertical  lines)  is  the  same  as  it  was  in  the 
circle,  and  if  we  call  the  vertical  force  on 
a  quadrant  V  (=  B  M)  for  the  circle  and 


FIG.  21. 
i 


i          1         1         i         JrjTT*    i        1         1        i 


^  M 


42 

V  (=:  B'  M')  for  the  ellipse,  we  will  have 
V  =  V  .  .  .  (21.) 
Notice,  however,  that  in  the  ellipse  the 
force  V  is  distributed  over  the  distance 
O'  B'  and  not  over  a  distance  =  O  B. 
Hence  the  intensity  of  the  force  V,  or 
the  amount  of  that  [  force  on  each  unit  of 
distance,  is  not  the  same  as  in  the  circle. 
In  the  ellipse  (Fig.  21)  each  little  verti- 
cal line  represents,  therefore,  the  force  on 
a  distance  greater  than  a  unit.  Let  O'  B' 
=  c  O  B.  Then  to  obtain  the  intensity 
of  V,  divide  it  by  the  space  over  which 
it  is  distributed.  Thus,  let 

V  H 

^-QB^^XO 

represent  the  vertical  and  horizontal  in- 
tensities in  the  circle.  We  have  al- 
ready seen  that  in  the  circle 

Py  =  Px  =  p. 

Let  py  and  p'x  represent  the  vertical 
and  horizontal  intensities  in  the  ellipse. 
Then 

«'   -    V  -      Y      =J^ 

py~O'  B'~c.OB        c  (22.) 


43 

The  lines  representing  the  "  pulls  "  at 
B  and  C  (as  B  N)  are  also  unchanged. 
Hence  the  pulls  at  those  points  in  the 
elliptic  ring  are  the  same  as  in  the  circu- 
lar ;  that  is,  they  are  equal  to  V  =  V. 

The  horizontal  lines  are  all  increased 
in  length  in  the  ratio  1 :  c.  Hence  the 
sum  of  the  lines  representing  the  hori- 
zontal force  on  a  quadrant  of  the  ellipse 
(as  I'  S')  is  greater  than  the  correspond- 
ing line  (I  S)  in  the  circle  in  the  above 
ratio.  Therefore  if  H'  ==  the  horizontal 
force  on  the  elliptical  quadrant, 

H'  =  c  .  H       .       .        (23.) 

The  length  over  which  this  force  H'  is 
distributed  (A'  O')  does  not  change, 
however,  and  hence  the  little  horizontal 
lines  in  both  figures  represent  the  force 
on  a  unit  of  distance.  Hence  the  inten- 
sity of  the  horizontal  force  in  the  ellipse 
has  increased  just  as  the  length  of  the 
lines,  or  from  the  equation 

H'         c.  H 
Px=&-0'=^0  =  Cpx      (24.) 

The  horizontal  pull  in  the  ring  at  A'  or 


44 

I'  being  equal  to  the  horizontal  force  on 
a  quadrant  is 

H'  =  <j.H  =  e.  V=  c.  V    .        .    (25.) 

Hence  the  "  pull "  around  the  ellipse 
is  not  constant  as  it  was  in  the  circle. 
The  pulls  at  B'  and  A'  are  as 

V  :  H'  :  :  i  :  c. 
But 

AT    :  C'B'  :  :  i  :  c. 

Therefore, 

1.  The  pulls  in  an  elliptical  ring  are 
as  the  axes  to  which  they  are  parallel. 

Again  the  intensities  in  the  ellipse  are 
p'y.p'x:  :  ~- :  cpx  :  :  --  :c:  :  i  :  c2 

c  c  , 

And 

(AT)2  :  (C'B')8  :  :  i  :  c* 

Therefore, 

2.  The  intensities  of   the  forces  in  an 
ellipse  are  as  the  squares  of  the  axes  to 
which  they  are  parallel. 

From  this  proportion  we  have 

c=J^-      .        .        .     (26.) 


45 

It  will  be  noted  in  the  elliptic  ring 
that  the  resultant  of  the  little  horizontal 
and  vertical  loads  at  any  point  is  not 
normal  to  the  curve  except  at  the  ex- 
tremities of  the  axes. 

Let  us  determine  the  pulls  and  the 
relations  between  the  forces  at  other 
points  besides  the  extremities  of  the 
vertical  and  horizontal  axes  of  the 
ellipse. 

In  the  circle  (Fig.  22)  if  we  resolve 
the  forces  along  any  two  rectangular  axes, 
as  A1  I,  and  C,  Bt,  we  shall  have  evidently 
the  same  relations  between  them  as  when 
resolved  along  a  vertical  and  horizon- 
tal axis.  Now  the  three  parallel  lines, 
viz.,  the  diameter,  A.l  Ip  and  the  tangents 
at  Cj  and  Bx,  are  projected  in  the  ellipse 
into  three  parallel  lines,  viz.:  A',,  I',,  and 
the  tangents  at  C',  and  B\.  Similarly 
Cj,  B,,  and  the  tangents  at  A1  and  Ij 
continue  parallel  in  the  ellipse.  Hence 
rectangular  diameters  of  the  circle  be- 
come conjugate  in  the  ellipse.  The  lines 
representing  the  forces  perpendicular  to 
C,  Bt  in  the  circle  become  parallel  to  O' 


46 
Fie.  22. 


47 

I\  in  the  ellipse,  and  are  changed  in 
length  just  as  O'  I\  is  changed  from  O  Ir 
So  the  forces  which  are  parallel  to  Cl  O 
in  the  circle  become  parallel  to  C\  O'  in 
the  ellipse,  and  vary  as  C\  O'  does  from 
0,0. 

Let  O'  !',=  /  and  O  0',=  /'  and  let 
the  total  force  parallel  to  O'  I',  on  a 
quadrant  (such  as  C\  I\  or  I',  B',)  of 
the  ellipse  be  =  V,  and  that  parallel  to 
O'  B/  be=H1.  Then  if  r  =  radius  of 
the  circle,  we  have  (since  the  force  on  a 
quadrant  of  the  circle  as  Cl  II  is  =  H  = 
V  =  T) 


H:Hl 


Hj  is  equal  to  the  pull  along  the  ring  at 
A't  or  I',  and  V,  is  that  at  0^  and  B',. 

JLeucQ  proposition  1  may  be  applied 
generally  to  all  conjugate  diameters  in 
the  ellipse  ;  that  is, 

3.  The  total  pulls  along  the  ring  at  the 
extremities  of  any  two  conjugate  diam- 


48 

eters  are  as  the  diameters  to  which 
they  are  parallel. 

Again,  the  intensities  being  equal  to 
the  total  loads  divided  by  the  surfaces 
over  which  they  are  distributed,  let 

p'y  =  intensity  of  load  parallel  to  O'  I\ 

P'x  "  "  "         C'  O' 

i 

Then 

Vi          V  r'  r1  "I 

y  —    f\i~rv~       ~~     //     j^y       // 
i       \j  \j  i         r.T  r 

p'  Hi  llr"  _          r" 

1~~O'  I/    ~  r  .  r"  ~PX    r' 

•    .     >  r    .        r"  .        r' 

i'       i  '    '     y  r"  r'          r' 

~^~  -  -  r  - 
Hence  for  proposition  2,  we  may  read, 
4.  The  intensities  of  a  pair  of  conju- 
gate loads  are  to  each  other  as  the 
squares  of  the  conjugate  diameters  to 
which  they  are  respectively  parallel. 

To   pass  from  one    set  of    conjugate 
forces  on  the  ellipse  to  another  ;  let 
p'x  and  p'y  be  the  intensities  parallel  to  one 

set  of  conjugate  diameters. 
HI  and  Vi  be  total  pulls  parallel  to  same  set 

of  conjugate  diameters. 
r"  r'  be  the  conjugate  semi-diameters. 


49 

Also  let 

i       i' 

be  the  corresponding  quantities  for  the 
other  set.     Then 


r"  ,     r 

P  X  =  PX  -    ,~  .    '  .  PX  —P  X Tf 

i            r  i  r 

r"\  ,             ,   r'r"i 

r\  l   i         | r\  r"  \ 
Also, 


TT  »»•     ~  TT  Hi    r  ' 

Hi  =  — —  '         orH=   — ,-       ^(29.) 
^ 

Similarly 


TT   /          Hr    '  •     TT'  ±Al  T   1 

HI    =-^T-        .    .H1=  ~^- 


Vi'=Vi  -V 

The  ellipse  (Figs.  21  and  22)  is  the 
form  assumed  by  a  cord  under  a  load 
composed  of  horizontal  and  vertical  com- 
ponents which  are  constant  along  the 
horizontal  and  vertical  lines,  but  which 
differ  from  each  other  in  intensity. 

The  diameter  C'  B'  of  the  ellipse  (Fig. 
21)  might  have  been  made  shorter 


50 

instead  of  longer  than  that  of  the  circle, 
if  required. 

Cor. — If  one  set  of  the  forces  are  ver- 
tical and  the  other  not  horizontal,  but 
inclined  at  an  angle  to  the  horizon  (Fig. 
23),  we  still  have  an  ellipse,  the  direc- 
tions of  the  forces  giving  the  directions 
of  two  conjugate  diameters  (A.\  O'  and 

B\  O').     Then,  if  p'x  =  the   intensity  of 

i 
the  inclined  force  and     ^—  intensity  of 

the  vertical  force,  we  have  by  proposition 

4, 

p'x  :p'v  :  :  (Bi '  O')2  :  (A!  '  O')2 

So  from  proposition  3,  if  V1  =  pull 
along  the  cord  at  B/  or  C/  and  H1  = 
that  at  A/ 

H!  :  Vi  :  :   BI  '  O'  :  A!  '  O'. 

From  the  first  of  these  propositions 
we  have  the  ratio  of  the  conjugate  dia- 
meters ;  and  from  the  second  we  find  the 
pulls  at  the  extremities  of  those  diam- 
eters. 

Knowing  two  conjugate  diameters  and 
the  angle  (90°-^*)  between  them  we  can 
readily  obtain  the  ellipse. 


51 

To  obtain  the  pulls  at  the  extremities 
of  any  diameter,  such  as  C1  B,. 

This  is  merely  passing  from  one  set 
of  conjugate  diameters  to  another  and 
FIG.  23. 


equation    (29)  gives  the  pull  at  B,,  for 

instance,  as 

O'K 


52 

(O'  K  being  conjugate  to  C,  O'  BJ,  etc., 
etc. 

An  important  fact  is  now  to  be  noted. 
Whenever  the  load  on  a  cord  is  entirely 
normal  to  it,  at  that  point  the  pull 
along  the  cord  is  equal  to  the  intensity 
of  the  normal  load  multiplied  by  the 
radius  of  curvature. 

For  the  cord  at  that  point  is  similarly 
situated  to  a  circular  cord  of  the  same 
curvature  and  under  a  load  of  the  same 
intensity. 

Thus,  in  the  ellipse  (Fig.  21)  the  action 
of  the  load  at  the  extremities  of  the 
axes  is  entirely  normal,  for  at  A'  and  I' 
the  horizontal  component  of  the  load 
vanishes  and  leaves  only  the  vertical, 
which,  at  these  points,  is  normal  to  the 
curve.  So  at  C'  and  B'  only  the  horizon- 
tal load  has  value,  and  its  action  is 
there  normal  to  the  curve. 

Consider  the  elementary  arc,  d  s,  at  A', 
for  instance,  which  is  subjected  to  this 
normal  load.  It  is  balanced  under  the 
equal  pulls  T  =  T'  (Fig.  24)  coming 
from  the  adjoining  parts  of  the  cord,  and 


53 

the  normal  load  p  d  s,  which  gives  it  its 
curvature.  Imagine  a  circle  under  a 
constant  normal  force  of  intensity  =  p. 
Take  an  equal  little  arc  d  s  of  it, 
loaded  with  a  normal  load  =  p  ds.  Then, 
if  it  be  acted  on  at  its  two  ends  by  ten- 
sions =  T  =  T',  it  is  evident  that  it  will 
have  the  same  curvature  as  the  arc  of  the 
ellipse  ;  or,  conversely,  if  it  has  the  same 
curvature,  the  pull  around  the  circle 
must  be  —  T=Tf. 

FIG.  24. 


Hence,  having  given  the  load  on  the 
curve  at  any  point  where  it  is  normal, 
we  determine  easily  the  pull  along  the 
cord  at  that  point.  For,  in  the  circle, 

H  =  V  =  T  =  px  r  =  py  r  =  p  r, 
and  in  the  ellipse  at  A' 

H'  =p'yp 


54 

Where  p  —  radius  of  curvature.  If  A'  O' 
=  r  and  O'  B'  =  c  r  in  the  ellipse  (Fig. 
21)  we  have  at  A' 


p  = =  ca  r. 

r 

.'.H'  =  p'  y.  c2  r  r=  -^-c2  r  — 
c 

So  in  the  parabola  under  uniform  ver- 
tical loads  (Case  I.)  we  have  seen  that  H 
=  2  p  m  (Bankine's  C.  E.,  p  165).  But 
H  =  p  p  ==  %  p  m  (since  p  =  2  m  at 
the  vertex). 

If  the  load  be  everywhere  normal  to 
the  cord  the  above  equation  will  apply  to 
every  point,  or 

T  =pp 

be  a  general  equation  of  the  curve. 

And  further,  when  the  load  is  every- 
where normal  we  have  already  seen  that 
the  pull  along  the  cord  must  be  constant, 
as  there  is  no  tangential  force  to  change 
it.  Hence, 

T  =  p  p  =  a  constant.  (31.) 

"When  the  load  p  is  constant,  of  course, 


55 


p  must  be  constant  too,  and  we  have  the 
circle  already  discussed.  When^?  varies, 
p  must  vary  inversely  as  p. 

FIG.  25. 


Case  VI.    Ifp  increases  in  value  just 
in   proportion    to    the   distance  of   the 


56 

points  of  the  cord  above  a  horizontal 
line  M  N  (Fig.  25),  the  cord  assumes 
the  shape  of  the  hydrostatic  arch.  This 
curve  possesses  geometrically  the  loops 
shown  in  the  figure  and  may  be  extend- 
ed indefinitely,  but  for  our  purpose  it  is 
evidently  only  necessary  to  discuss  that 
part  between  the  points  C  and  B  (Fig. 
26)  where  the  tangents  are  vertical. 

Taking  L  (Fig.  26)  for  the  origin,  if  the 
intensity  of  the  load  then  be  y Q  (=.  A. 
L).  multiplied  by  a  constant,    or   w  y0, 
then  at  any  other  point  it  is  =  w  y. 
Hence  the  equation  of  the  curve  is 
T=pp=wyp=wy0p0  =  a  constant 

(y0  and  pQ  are  the  values  of  the  ordinate 
and  radius  of  curvature  at  A). 

Let  us  resolve  the  normal  load  on  C 
A  B  as  we  did  in  the  circle,  into  its 
horizontal  and  vertical  components.  As 
was  the  case  in  the  circle,  these  will  be 
for  each  point  equal  in  intensity  to  each 
other  and  also  to  the  normal  force,  or 

p  =  px  —  pv- 
But   these   quantities   are   no   longer 


58 

constant  (as  in  the  circle)  all  along  the 
curve,  but  vary  from  point  to  point. 

If  we  form  the  parallelogram  of  forces 
for  any  arc  A  D  (as  in  Fig.  27),t.he  side 
N  F  =  F  S,  since  H  =  T  =  a  constant, 
and  F  G  must  represent  the  resultant  of 
the  whole  load  on  A  D  both  in  amount 
and  direction. 

The  vertical  component  F  E  of  F  G  is 
equal  to  the  vertical  component  S  X  of 
SF,  or 

Vertical  load  on  A  D  =  T  sin  i  =  H 
sin  i. 

At  B  the  vertical  load  =  T  =  H  =  Y 
(since  i  =  90°  there). 

So  the  horizontal  component  of  the 
total  load  on  A  D  is  G  E,  and  since 

NF  =  GS  =  GE  +  FX 
we  have  horizontal  load  on 

A  D  =  G  E  =  N  F-F  X  =  H-H  cos  i  = 
H  (1— cos  i). 

At  B,  i  =  90°  .-. 
Horizontal  load  on  A  B  =  H 
On  the  arc  D  B 

Horizontal  load  =  H— H  (I—cost)  =  H  cos  i. 


59 

The  vertical  load  on  A  D  may  be  thus 
expressed  : 


H  sin  i  = 


Q  o0  sinj 


FIG.  27. 


The  horizontal  load  thus  : 


60 
wf  ydy=w.  y  ~y°  (33.) 

And  if  yi  =  ordinate  of  B,  the  hori- 
zontal load  on  A  B  is 

H  =  wy^~^°2  (34.) 

For  formula  for  radius  of  curvature 
see  Bankine,  C.  E 

The  equation  T  =  H  =  w>2/0  p0  =  w 
y  p,  enables  us  to  solve  problems  simi- 
lar to  those  under  the  parabola. 

Case  VII.  If  we  construct  a  curve 
from  the  last  one  by  using  the  same  or- 
dinates  and  by  changing  all  the  abscissas 
in  the  ratio  c :  1,  so  that  the  new  co-or- 
dinates of  a  point  shall  be  y  and  c  x, 
and  at  the  same  time  change  the  hori- 
zontal forces  in  the  same  proportion, 
leaving  the  vertical  ones  unchanged, 
the  new  curve  and  new  system  of  forces 
so  obtained  will  evidently  be  parallel 
projections  of  the  former,  and  will  be 
balanced.  This  new  curve  C'  A  B' 
(Fig.  28)  is  the  "  Geostatic,"  and  bears 
a  relation  to  the  "  Hydrostatic  "  strict- 


61 

ly  analogous   to    that  between    the    el- 
lipse and  circle.        Hence, 

The  total  vertical  load  on  A  B'  =  V  —  1 

V  =  pull  along  cord  at  B'.  l,q~  N 

Total  horizontal  load  on  A  B'  =  H'  =  f  ^OJ 
c  H  =  pull  along  cord  at  A'.  J 

FIG.  28. 


62 
The  intensities  are 


H'        £   H 

For  horizontal  load  p'x  =rrr  =TTT  =^?o;  I 

DA.      O  A. 

(V  H  Pa.  and  jt?y  referring  to  the  hy- 
drostatic curve.) 

The  load  at  A  and  B'  and  C'  being  al- 
together normal  (it  is  not  so  at  the 
other  points),  let 

p'0  and  p\  be  the  radii  of  curvature  at 
A  and  B'. 

Then 


In  the  hydrostatic 
H=  Pyp0.         .'.  cH  —  R'  =  cpyp0. 

Py     < 
•'•  Po  = 


P'O  =  0«  Po        •         • 

So 

V  =p'xp'i  =cpxp'i    =  V. 

But  in  the  hydrostatic 


•'•PxPi  = 

.  •'    - 


(38.) 


63 


These  radii  are  useful  in  drawing  the 
geostatic  curve. 

Case  VIII.  So  far  we  have  dis- 
cussed the  curves  assumed  by  cords  un- 
der loads  distributed  according  to  some 
simple  law.  But  it  is  possible  to  dis- 

FIG.  29. 


cuss  the  more  general  problem  :  Given 
a  load  that  varies  and  is  distributed  in 
any  manner,  required  the  curve  which  it 
will  cause  the  cord  to  take  ;  or  converse- 
ly, given  a  curve,  required  the  character 
and  distribution  of  the  load  to  produce 
it.  The  most  useful  form  of  the  prob- 
lem is  that  in  which  we  assume  the 


64 

shape  of  the  cord,  and  the  vertical  com- 
ponents of  the  load,  and  require  to  be 
found  the  intensity  and  distribution  of 
the  horizontal  components  of  the  load 
necessary  to  produce  equilibrium. 

To  illustrate :  Assume  the  curve  to  be 
a  circle,  and  the  vertical  load  to  be  uni- 
form in  intensity,  we  see  at  once  that 
the  horizontal  load  should  be  also  uni- 
form, and  of  intensity  equal  to  that  of 
the  vertical  load. 

But  generally  :  Let  C  A  B  (Fig.  29) 
be  some  assumed  curve,  and  let  the  ver- 
tical load  be  known  in  amount  and  dis- 
tribution. Making  some  changes  in  the 
signification  of  the  letters  heretofore 
used,  now  let 

Y  =  vertical  load  on  any  arc  A  D. 

Y!  —  vertical  load  on  the  semi-cord  A  B. 

H  =  horizontal  load  on  any  arc  A  D. 

H!  =  "  half -cord  A  B. 

H0  =  pull  along  cord  at  A  (the  quantity 

heretofore  denoted  by  H). 
Px  and  py  —  the  horizontal   and  vertical 

intensities  as  heretofore. 
PQ  =  value  of  py  at  the  point  A. 
P0  and  PI  =  radii  of  curvature  at  A  and  B. 


65 
The  vertical  load  on  an  arc  A  D  is 


V  =         pydx        .        .     (39.) 
•/   o 

Again  at  the  horizontal  point  A,  the 
vertical  projection  of  the  element  of  the 
curve  being  =  zero,  the  load  is  entirely 
vertical,  and  consequently  at  that  point 
is  normal  to  the  curve.  Hence  the  pull 
along  the  cord  at  A  is 

H0  =PoP0. 

To  discuss  the  forces  upon  an  arc  A 
D.  Draw  tangents  at  A  and  D.  They 
meet  at  F  (Fig.  29),  through  which  point 
the  resultant  of  the  total  load  on  A  D 
must  pass.  The  vertical  load  is  also  = 
the  vertical  component  of  the  pull  along 
the  cord  at  D,  for  these  two  forces,  be- 
ing the  only  vertical  ones  connected 
with  A  D,  must  needs  balance  each 
other,  therefore, 

Lay  off  F  N  =  H0.     Lay  off  F  E  ver- 

tical and  ~y    pydx.  Complete  the  rect- 

angle F   E  S  X.      The  pull  along  the 
cord  at 
D  =  F  S  =  F  E  cosec  i  =  V  cosec  i  .  .  (40.) 


66 


Also, 

S  E  =  F  X  =  V  cot  i  =  horizontal  compound 
of  pull  along  the  cord  at  D      .       .      (41.) 

But  the  horizontal  pull  at  A  is 

H0  =FN  =  GS. 

.-.  G  E  =  H0  -  V  cot  =  H  =  resultant 
of  horizontal  load  on  A  D  (42.) 

The  intensity  of  this  horizontal  load 
may  be  expressed  thus  : 


_d  H  _        6  (V  cot  i)  _       \ 
PX~~  dy   '-  dy  dy      (43.) 

At  B  the  vertical  load  =  Vt.  Let 
this  be  represented  by  B  K  (Fig.  29). 
If  the  cord  be  itself  vertical  at  that 
point,  B  K  —  .  Vj  will  be  equal  to  the 
pull  along  it  at  B.  If  the  cord  is  in- 
clined as  in  the  figure,  draw  its  tangent 
atB,  and 

B  L  —  B  K  cosec  e\  =  Vi  cosec  it  =  pull 

along  the  cord. 
and 

K  L  =  B  K  cot  ii  •=  Vj  cot  i±  =  horizontal 

component  of  this  pull. 
H0  —  Vj  cot  il   =  H!  =  resultant  of  entire 

horizontal  load  on  A  B. 


67 


It  may  often  happen  that  S  E  =  V  cot 
i  =  horizontal  component  of  the  pull 
along  the  cord  at  D  (Fig.  30)  is  greater 
than  GS  =  FN  =  H0  =  horizontal 
pull  along  the  cord  at  A.  In  such  cases 
G  E  —  H0  —  Y  cot  i  is  negative,  which 
indicates  that  the  horizontal  load  be- 
tween A  and  D,  for  at  least  a  part  of 

FIG.  30. 


this  distance,  must  be  contrary  in  direc- 
tion to  that  heretofore  discussed  ;  that  it 
must  exert  an  inward  pull  instead  of  an 
outward  one  (Fig.  30).  If  this  "inward 
pull "  were  removed  or  replaced  by  an 
outward  one,  the  curve  would  evidently 
be  flattened  about  A. 


68 

We  may  illustrate  geometrically  the 
relation  between  the  forces  in  all  parts 
of  AB. 

The  vertical  load  and  curve  being- 
given,  draw  F  Ev  (Fig.  32)  =  the  fatal 
vertical  load  on  A  B,  and  lay  off  on  it 

F  E'  =  vertical  load  on  the  arc  A  D'. 
F  E"  =       "          «        «  A  D",  etc. 


Draw  a  horizontal  line  at  F  and  lay 
off  F  N  and  F  K,  each  =  H0  =  pull  at 
A.  Draw  through  F  lines  parallel  to  the 


69 

tangents  at  D'  D"  D"',  etc.,  and  through 
E'  E"  E'",  etc.,  lines  parallel  to  the 
horizon.  Then  the  oblique  lines  F  S',  F 
S",  etc.,  represent  the  pulls  along  the 
cord  at  V  D",  etc.,  while  E'  S',  E"  S', 
etc.,  represent  the  horizontal  components 
of  these  pulls.  Lay  off  from  each  point 
S'  S",  etc.,  horizontal  lines,  each  equal  to 
F  N,  and  draw  through  the  points  G' 
G",  etc.,  thus  obtained,  a  curve.  It 
will  evidently  be  similar  to  that  drawn 
through  K  S'  S",  etc.,  and  the  line  G'  E' 
will  represent  the  resultant  of  the  hori- 
zontal load  that  must  be  distributed 
along  the  curve  from  A  to  D';  G"  E", 
the  resultant  of  the  horizontal  load  be- 
tween A  and  D",  and  so  on. 

(Fig.  32)  is  really  formed  from  the 
parallelogram  of  forces  for  the  arcs  A 
D',  etc.;  this  parallelogram  being  at  D'  = 
F  N  G'  S',  in  which  E'  S'  is  the  horizon- 
tal component  of  the  pull  at  D  and  W 
G'  =  the  resultant  of  the  horizontal 
load  on  A  D'. 

As  the  abscissas  of  the  curve  F  G'  G"r 
etc.,  increase  to  the  left  of  F  Ev  from 


70 


the^point  F  to  G''  (which  correspond  to 
D"  on   the   curve),  the    horizontal  load 

FIG.  3 1 . 


acts  outward  011  the  arc  A  D".     The  ab- 
scissas  then  diminish    to    G'".      Hence 


71 

between  D"  and  D'"  on  the  curve,  the 
horizontal  load  must  act  inwards  as 
shown  in  (Fig.  31).  From  Of"  the 
abscissas  increase  until  we  reach  Gv. 
Hence  the  horizontal  load  acts  outward 
throughout  the  remainder  of  the  cord. 
The  points  n  and  n'  correspond  to  those 
arcs  on  which  the  resultant  of  the  hor- 
izontal load  is  zero.  Thus  on  the  arc  A 
n  the  negative  horizontal  load  is  just 
equal  to  the  positive,  and  hence  their 
sum  —  zero.  So  on  the  arc  A  n'. 

Note  that  the  abscissas  of  the  curve 
F  G'  .  .  .  Gv  are  not  the  intensities  of 
the  horizontal  loading,  but  that  each  such 
abscissa  represents  the  algebraic  sum  of 
the  entire  horizontal  load  between  A  and 
the  point  to  which  the  abscissa  corre- 
sponds. The  intensity  in  question  has 
already  been  shown  to  be 

dB. 


dy 


In  this  expression  d  H  =  the  differ- 
ence of  two  neighboring  abscissas  of  the 
curve  F  G'  .  .  .Gv;  as  for  instance,  d  H 


72 

—  G'  E'  —  G"  E".  And  d  ?/  =  vertical 
projection  of  the  arc  D'  D''  of  the  cord 
to  which  the  above  corresponds. 

ARCHES. 

FIG.  33. 


ii 


Let  us  imagine  the  curve  of  the  cord 
to  be  reversed,  and  the  cord  itself  to  be 
replaced  by  a  thin  metal  strip,  which, 
like  the  cord,  shall  be  practically  without 
transverse  stiffness,  but,  unlike  the  cord, 
shall  be  able  to  resist  a  compressive 
force  in  the  direction  of  its  length  at 
every  point.  Let  the  loads  be  distrib- 
uted as  heretofore,  except  that  where 
there  are  horizontal  components  of  the 
load,  these  should  act  inward,  where 
upon  the  cord  they  acted  outward,  and 
vice  versa.  We  then  have  what  is  called 


73 

a  "  linear  arch  or  rib  " ;  and  the  curve 
assumed  by  it  will  be  identical  with  that 
of  the  cord  under  equal  and  similarly 
distributed  loads.  If  the  loading  is 
changed  in  distribution,  the  rib  will 
change  in  shape  just  as  the  cord  would 
do  under  similar  circumstances. 

In  practice  there  are  no  "linear 
arches,"  but  the  discussion  of  them 
enables  us  to  determine  the  form  of 
equilibrium  for  real  arches.  If  we  know 
the  form  that  a  linear  arch  would  as- 
sume under  a  given  load,  we  can  find 
the  "line  of  pressures"  in  the  real  arch. 
This  line  and  the  value  of  the  thrusts 
at  all  its  points  enable  us  to  solve  the 
problems  that  arise  in  arch  building. 

1.  Suppose,  for  instance,  we  desire  to 
construct  an  arch  to  bear  a  uniform  ver- 
tical load,  such  as  that  discussed  in  Case 
I.  The  shape  of  the  linear  rib  under 
such  a  load  is  a  parabola.  We  then  have, 

1°  Step.  Assume  this  curve  for  the  in- 
trados  CAB  (Fig,  34).  If  the  arch  and 
the  load  be  of  homogeneous  material, 
the  shape  of  the  extrados,  or  outside  of 


74 


the  load,  will  be  M  Y  M',  the  vertical 
distance  between  C  A  and  M  Y  being 
constant. 

2°  Step.  Is  to  determine  the  depth 
A  L  of  the  keystone.  This  depth  is  al- 
ways greater  than  necessary  simply  to 
prevent  the  crushing  of  the  material  of 

FIG.  34. 


the  arch  under  the  thrust  at  the  crown. 
Prof.  Eankine's  empirical  rule  derived 
from  the  best  examples  is  to  make  the 
depth  of  the  keystone  in  feet 

In  single  arches  1 

"  V.  12  x  radius  of  curva'e  at  the  crown.  !  ,,/,  v 

In  arches  of  a  series 
=   V.17  x  radius  of  curva'e  at  the  crown.  J 


75 

3°  Step.  Determine  whether  the  "  line 
of  pressures"  can  He  in  the  "middle 
third  "  of  the  ring  of  voussoirs.  It  should 
be  restricted  to  the  middle  third  to  pre- 
vent the  voussoirs  tending  to  open  at 
any  of  the  joints. 

We  can  test  this  as  follows  : 

Suppose  the  voussoirs  to  be  constant 
in  depth  all  around  the  arch  as  in  (Fig. 
34).  Consider  any  part  of  the  arch  in- 
cluded between  the  vertical  plane  (A  L) 
at  the  crown,  and  a  vertical  plane  at  any 
other  point,  as  D'  P'.  The  calculated 
horizontal  thrust  along  the  linear  rib, 
which  coincides  in  shape  with  the  soffit 
C  A,  is  indicated  by  the  arrow  with  its 
head  at  A.  Let  the  horizontal  thrust 
of  the  rib  at  D'  be  indicated  by  the  ar- 
row with  its  head  at  D'  pointing  in  an 
opposite  direction  to  that  at  A.  At  the 
crown  take  A  K  not  greater  than  %  A 
L.  Imagine  a  left-handed  couple  ap- 
plied to  A  L  in  the  vertical  plane  of  the 
arch,  whose  force  =  H  =  the  thrust  at 
A,  and  whose  lever- arm  =  A  K.  Apply 
an  equal  and  opposite  couple  on  the  plane 


76 

D'  P',  with  a  force  H',  equal  to  the  hori- 
zontal thrust  of  the  rib  at  D'.  Its  lever- 
arm  Dr  P'  must  then 

H.  AK 


H' 

In;  the  parabola  H  =  H'  .-.  D'  P'  =  A 
K.  These  couples  being  equal  and  op- 
posite do  not  change  the  conditions  of 
equilibrium  of  the  section  of  the  arch  L 
D',  but  they  transfer  the  line  in  which 
the  thrust  acts  from  A  D'  to  K  P.  We 
can  repeat  the  process  as  often  as  we 
choose  by  taking  parts  L  D",  etc.;  and 
if  the  curve  drawn  through  the  points 
K  P'  P",  etc.,  lies  within  the  middle 
third  of  the  arch-ring,  the  arch  is  suffi- 
ciently stable. 

In  the  case  before  us,  the  horizontal 
thrust  being  constant  for  every  point  of 
the  rib  C  A,  the  lever-arms  D'  P',  D"  P", 
etc.,  are  also  equal,  and  therefore  the 
"  line  of  pressures  "  K  K'  is  merely  the 
parabola  raised  vertically  a  distance  =  A 
K.  If  K  K'  does  not  lie  in  the  middle 
third,  a  slight  increase  in  the  voussoirs, 


77 


especially   towards   the   springing,    will 
usually  remove  the  difficulty. 

4°  Step.  The  joints  between  the  vous- 
soirs,  such  as  D'  G  (Fig.  35)  are  usually 
made  normal  to  the  soffit  A  C,  but  whether 
this  be  done  or  not,  the  direction  of  G 
D'  must  be  such  that  at  S,  where  the 


line  of  pressures  cuts  it,  the  angle  in- 
cluded between  S  N  (the  normal  to  G  D') 
and  S  T  (the  tangent  at  S  to  K  K')  may 
be  less  than  the  angle  of  friction  of  the 
material  of  the  voussoirs.  The  best  pos- 
sible direction  for  the  joints  D'  G,  etc., 


78 


would  be  to  make  them  perpendicular  to 
KK'. 

The  horizontal  component  of  the 
thrust  (H)  along  the  curve  of  pressures 
in  a  parabolic  arch,  is,  as  we  have  seen, 
constant ;  but  the  thrust  along  the  curve 
(T)  increases  from  A  to  C,  and  its  value 
at  any  point  may  be  determined  by  the 
formulae  in  Case  I. 

Parabolic  stone  or  brick  arches  are  noi 
common,  because  it  is  rare  to  have  such 
a  distribution  of  the  load  as  that  sup- 
posed above. 

2.  But  if  we  reverse  the  curves  dis- 
cussed under  Cases  II.  and  III.,  we  have 
a  form  of  arch  much  more  frequently  ap- 
plicable. 

Thus,  suppose  the  arch  and  its  back- 
ing to  be  homogeneous,  and  that  the 
extrados  of  this  loading  is  horizontal 
(M  Y),  and  suppose  the  action  of  the 
load  to  be  entirely  vertical.  Then  the 
arch  and  its  backing  are  similar  to  the 
metal  sheet  and  the  cord  discussed  in 
the  cases  just  referred  to,  and  therefore 
the  form  of  the  linear  rib  under  such  a 


79 

load  will   be  a  catenary  or   transformed 
catenary— usually  the  latter. 
FIG.  36. 


Assume  this  curve   for  the  soffit  C  A 


80 

B  (Fig.  36) ;  determine  the  depth  A  L  ; 
the  line  of  the  pressures  K  K';  and  the 

FIG.  37. 


direction  of    the  joints  ;    as  in  the  last 
case.     In  this  case,  as  in  the  parabola, 


81 


H  is  constant,  and  hence  the  curve  of 
pressures  is  merely  the  curve  C  A 
raised  vertically  through  a  distance  = 
A  K. 

Example. — Let  the  data  for  a  required 
arch  be  (Fig.  37)  span  C  B  =  10' ;  rise 
O  A  =  4';  height  of  extrados*  M  Y 
above  springing  at  C  =  10'.  Let  the 
arch  and  brickwork  be  of  solid  brick- 
work whose  weight  w  per  cubic  foot  = 
112  Ibs. 

The  equation  of  the  transformed  cate- 
nary passing  this  C  A  B  is 


Where  y0  =  A  Y  —  6'  (the  origin  be- 
ing at  Y  and  the  axis  of  abscissas  hori- 
zontal). 

First  find  m,  the  modulus  of  the  cor- 
responding common  catenary.  By  eq. 

(14) 


V 


82 


At  the   point  C  xr  =  5   ft.  and  y'  = 

10ft. 

.-.  m  =  4.54  ft.  =  YN. 

Then  determine  points  of  the  curve, 
thus  ;  for 

x  =  1,  y  —     for  x  =  2,  y  =    for  #  =  3,  #  — 
for  x  =  4,  y  =     etc 

Describe  the  curve  through  these 
points. 

The  thrust  at  the  crown  A  is  (for  a 
unit  of  length  of  the  arch) 

H  =  wm*  from  eq.  (16). 
.-.  H  =  (112)  (4.54)8  =  2,308.3  Ibs. 
From  eq.  (15)  area  A  Y  M  C  = 


E™  -  E~  m      =  36.32  sq.  ft. 

Weight  of  load  A  Y  M  C  =  P  =  (112)  (36.32) 

=  4,067.84  Ibs. 

From  eq.  (18)  thrust  at  C  =  T  =  4/p«^-H2  = 
4,677.1  Ibs. 

Inclination  at  C  .  Tan  i,  =  *^±=^ 
dx,       2m 

\      xl          —*-  ) 
\    EW  —  E     m   )    : 

.-.  »\  =  60°  33'. 


83 


The  formula  for  depth  of  keystone  will 
be  satisfied  by  making  the  depth  of  the 
arch  A  L  =  length  of  one  brick  =  9", 


oc 

CO 


for  this  gives  9"  X  12"  =108  square 
inches  to  bear  the  thrust  H  =  2,308.3 
Ibs.,  or  T  =  4,677.1  Ibs.  The  latter  is 
the  greatest  thrust  in  the  arch. 


84 

It  is  easy  to  see  that  K  K'  will  be  in 
the  middle  third,  for  even  at  C  the  dis- 
tance of  the  point  of  the  curve  K  K' 
vertically  over  C,  from  the  nearest  point 
of  C  A,  is  approximately 

6  X  cos  (90°  -  60°  32')  =  5"+. 

The  extrados  of  the  transformed  cat- 
enary need  not  be  the  directrix  M  Y  ; 
it  may  be  another  transformed  catenary, 
provided  these  catenaries  have  the  same 
directrix. 

To  illustrate:  Suppose  the  weight  of 
a  unit  of  the  material  between  CAB  and 
M  Y  —  w.  Then  the  intensity  of  verti- 
cal pressure  at  any  point  G  of  CAB 
(Fig.  38)  is  =  w  y.  If  a  heavier  build- 
ing material  were  used  this  vertical  press- 
ure could  be  brought  upon  G  by  a  less 
height  of  it.  Let  this  heavier  material 
have  a  weight  per  unit  =  w  and  let 

2 
w  =—w. 

Then  a  column  of  the  heavier  material 
over  G  and  of  a  height  =  f  y  would  give 


85 


the  same  pressure  as  the  whole  column  of 
the  lighter,  or 


o 

--w'y        .        .      (47.) 


At  each  point  CAB  (Fig.  38)  lay  off 
two-thirds  of  the  vertical  ordinate,  and 
through  these  points  draw  C"  A"  B." 
The  upper  service  of  the  load  may  have 
this  form,  and  yet  CAB  still  be  the 
shape  of  the  linear  arch  balanced  under 
the  applied  forces.  The  equation  of  C 
A  B  being 


y  =••  -T 1  E™  +  E 

that  of  C''  A"  B"  is  evidently 


**> 


E    "I  (48.) 


The  principle  of  this  example  is  gen- 
eral. 

When  the  extrados  is  a  transformed 
catenary,  note  that,  since  in  all  the  for- 
mulae under  Case  III.,  w  =  the  weight 
corresponding  to  a  unit  of  surface  of 


86 

the  space  between  CAB  and  M  Y,  we 
must  make  in  these  formulae 

w  =  n  w' 

Where  w'  =   weight    of    the    building 
AA  " 


material  and  n  =• 


Y  A 


In  arches  of  this  class  no  provision 
is  needed  for  horizontal  thrust  on  the 
spandrels,  as  the  arch  is  equilibrated 
under  vertical  loads  alone. 

In  all  stone  or  brick  arches,  the 
changes  in  the  curve  of  pressures  K  K' 
due  to  passing  loads  are  usually  slight, 
because  the  weight  of  such  passing  loads 
is  generally  small  compared  with  the 
weight  of  the  arch  itself  and  its  backing. 

3.  The  simplest  practical  case  in 
which  a  uniform  normal  load  (such  as 
that  discussed  in  Case  IV.)  can  be  ap- 
plied to  an  arch  is  when  it  is  subjected 
to  water  pressure,  the  arch  ring  being 
horizontal  instead  of  vertical.  Such  a 
pressure  will  exist  on  an  empty  well  con- 
structed in  a  reservoir  or  other  body  of 


87 

water  (Fig.  39).  For  each  horizontal 
layer  of  the  well  wall  may  be  considered 
as  subjected  to  a  uniform  normal  press- 

FIG.  39. 


ure  of  an  intensity  due  to  the  depth  of 
the  water  at  that  layer.  This  intensity 
will  of  course  diminish  (and  so  will  the 


.    88 

pressure  on  the  wall)  from  one  layer  to 
another  as  we  come  towards  the  top. 

The  soffit  of  such  a  well  should  be  of 
circular  form  (Case  IV).  The  thickness  of 
the  wall  at  any  depth  must  be  determined 
by  the  thrust,  which  is  constant  all 
around  any  given  layer,  and  is 

T  =  p  r  =  w  y  r.        .        .     (49.) 

Where  w  =  weight  of  a  unit  of  water 
and  y  =  depth  of  water  at  the  layer  in 
question. 

In  determining  the  line  of  pressures 
consider  a  section  of  the  wall  befween 
two  yertical  planes  not  parallel  as  here- 
tofore, but  both  normal  to  the  soffit. 
Take  for  the  lever-arm  of  the  couple  at 
A  (Fig.  39)  a  distance  A  K  =  \  A  L. 
The  force  is  still  to  be  =  H  =  T. 

At  D  apply  an  equal  couple  with  force 
=  the  thrust  along  the  soffit  at  that 
point,  which  is  also  =  T  =  H.  Then  the 
lever-arm  must  be  equal  to  A  K.  Hence 
we  see  that  the  curve  of  pressures  is  a 
circle  parallel  to  the  soffit,  and  may  pass 
through  the  middle  of  the  arch  ring. 


89 


This  kind  of  arch  may  be  used  for 
dams  or  the  walls  of  reservoirs.  (See 
Fig.  40.) 


4.  There  is  no  case  in  ordinary  prac- 
tice where  the  pressures  upon  an  arch 
are  strictly  identical  with  those  on  an  el- 


90 


liptical  cord,  for  in  this  case  the  press- 
ure must  be  constant  in  intensity  along 
both  the  horizontal  and  vertical  projec- 
tions of  the  arch,  but  the  intensity  along 
the  horizontal  must  differ  from  that 
along  the  vertical  in  a  constant  ratio 
(Fig.  41).  But,  as  Prof.  Rankine  says, 

FIG.  41. 


H,  \  1  1  1  j 


the  curve  of  equilibrium  for  the  arch  of 
a  tunnel  through  earth,  when  the  depth 
below  the  surface  is  great  compared  with 
the  rise  of  the  arch  itself,  approximates 
to  an  ellipse. 

The  pressures  in   a  mass  of  earth  are 
intermediate  in  character  between  those 


91 


existing  in  a  solid  and  those  in  a  liquid 
mass.  Thus  a  little  cube  of  earth  (Fig. 
42)  under  the  weight  of  the  superincum- 
bent column  of  earth  /?,  presses  down- 

FIG.  42. 


J 


ward  with  a  force  equal  to  its  own 
weight  and  that  of  the  column  above. 
It  also  presses  out  horizontally  with 
a  force  less  than  this  downward  force, 


92 


but  always  bearing  a  constant  ratio 
to  it.  If  the  little  cube  were  solid  it 
would  have  no  horizontal  push  ;  if  liquid, 

FIG.  43. 

xL  Y 


that  horizontal  push  would  equal  its 
pressure  downward.  If  the  upper  sur- 
face of  the  earth  is  inclined,  the  outward 


93 

push  which  always  remains  parallel  to 
it  becomes  inclined  too,  and  is  then  "con- 
jugate "  to  the  vertical. 

If  M  Y  (Fig.  43)  is  the  surface  of  the 
earth,  when  Y  A  is  great  compared  with 
A  O,  then  Y  A  and  M  C  differ  so  slightly 
that  we  may  assume  them  to  be  equal. 
We  then  have  on  the  arch  a  uniform  ver- 
tical load  whose  intensity  = 

Py  =  (Y  A)  x  weight  of  a  unit  of  the  earth=w#0, 

and  a  horizontal  load  whose  uniform  in- 
tensity px  is  equal  to  the  vertical  inten- 
sity (py  )  multiplied  by  a  constant.  Let 

f)r 

-    =  c9  (a  constant). 
Py 
Then 

px  =  c*wy0  and  c  —  \  ^L  . 
py 

From  the  discussion  of  Case  V.  we  see 
that  c  must  be  the  ratio  of  the  axes  of 
the  ellipse  to  which  the  pressures  are  re- 
spectively parallel.  Hence  if  the  arch  be 
a  semi-ellipse  and  O  B  be  given,  we  have 

OB  OB 

?— r^  =  c  .'.  O  A  =— 
O  A  c 


94 

From  these  data  draw  the  curve  of  the 
soffit. 

The  thrust  along  the  soffit  at  A 

=  H  =  Py  9o   =  ^y0$V 
AtC  or  Bit  is  V  =  pxgim 
At  other  points  it  may  be   gotten  from 
eq.  (27)  Case  V. 

We  can  determine  the  curve  of  press- 

FIG.  44. 

JL 


ures  by  a  method  similar  to  that  used 
in  the  last  case.  Here,  however,  the 
curve  K  K'  will  not  be  parallel  to  C  A, 
since  the  thrusts  along  C  A  are  not  con- 
stant, but  increase  from  A  to  C.  Assume 


95 

A  K  (Fig.  44)  =i  f  A  L,  then  the  arch 
must  be  so  proportioned  that  K  K'  shall 
fall  within  the  middle  third. 

If  the  arch  C  A  B  is  not  to  be  a  ^mi- 
ellipse  (as  above  assumed)  but  only  a  seg- 
nient  of  one,  a  few  trials  will  enable  us 
to  get  the  ellipse  from  the  data  already 
given. 

The  strictly  true  curve  of  equilibrium 
required  by  earth  pressure  is  the  Geo- 
static  arch. 

5.  An  arch  built  with  the  curve  dis- 
cussed in  Case  VI.  is  known  as  the  Hy- 
drostatic arch,  from  the  fact  that  the 
loading  there  described  is  similar  to 
the  pressure  of  the  water  upon  a  vertical 
arch. 

For  if  M  Y  (Fig.  45)  be  the  surface 
of  the  water,  then  its  pressure  on  C  A  B 
is  normal  and  proportioned  at  each  point 
to  the  depth  below  M  Y.  This  press- 
ure, as  has  been  shown,  may  be  resolved 
into  a  vertical  and  horizontal  pressure 
at  each  point,  this  vertical  and  horizon- 
tal pressure  being  equal  in  intensity  to 
each  other  at  every  point,  and  also  to 


96 

the  normal   pressure  of   which  they  are 
the  components. 


to 
^ 

I 


The  above  form  of  arch  may  be  applied 
in  two  cases. 


97 

(1)  To  bear  the  pressure  of  water  or 
other  liquid.  Thus  in  the  case  of  a  river 
tunnel  (such  as  those  at  Chicago)  where 
the  top  of  the  tunnel  is  practically  on  a 
level  with  the  bottom  of  the  river,  we 
might  use  the  hydrostatic  arch. 

The  equation  of  the  curve  is 


The  vertical  load  on  the  half  -arch  A  B 
—   I    pdx=\=wy}  pl=thrust  along  arch  at  B. 

*      Xi 

The  horizontal  pressure  against  A  B 

y^-y<f 
y=iD  -  s  —  -  =  H  =  w#0/>o    .   (51. 


y\ 

The 'thrust  along  the  arch  is  constant,  or 
T  =  H  =  V. 

The  rise  A  O  (=  a),  the  depth  AY 
(=  f/0),  and  the  radii  at  A  and  B  (p0  and 
p,)  are  connected  by  the  following  ap- 
proximate equations.  The  co-ordinates 
of  B  being  xl  and  yl?  let 


(52) 


98 


Po 


=  V'  a,f°  =  a  +^7  =  in1*-^)       (53-) 


The  line  of   pressures  in  a  hydrostatic 


arch,  since  T  is  constant,  is  parallel  to 
the  soffit,  as  in  circular  arches. 


99 

Example.  —  Suppose  the  span  to  be  50 
ft.  (Fig.  46)  and  the  depth  A  Y  =  16  ft. 

Find  first  the  rise  A  O.  In  eq.  (52)  x1 
=  25'  y0  =  16',  and  a  few  trials  show 
that  a  =  rise  =  20'  about. 

Hence 

p0  =  32Jft.  and^  =  14.1  ft. 

With  these  data  describe  the  curve  of 
the  soffit  —  the  radius  at  any  other  point 
besides  A  and  B  being  given  by  the  equa- 
tion 


The  thrust  at  A  =  H  =  wyQ  $>0.     Here 
w  =  62.4  Ibs. 

.  .  H  =  32,448  Ibs. 

The  rule  for  the  depth  of  keystone  in  a 
single  arch  gives 

Depth  A  L  =  V  .12  x  32T5  =  1-9  ft- 

This  is  ample.  It  only  gives  about 
120  Ibs.  per  sq.  in.  as  the  pressure  at  the 
crown. 

T  being  —  H,  the  depth  of  the  arch- 
ring  may  be  uniform. 


100 

(2)  The  hydrostatic  arch  is  also  used 
when  the  loading  is  homogeneous  ma- 
sonry up  to  the  extrados  M  Y,  provided 
the  spandrels  be  suited  to  sustain  a 
horizontal  thrust  at  each  point  of  the 
arch  equal  to  the  vertical  load  at  that 
point. 

As  all  stone  or  brick  arches  sink  at  the 
crown  when  the  centers  are  removed, 
they  will  exert  at  other  points  an  out- 
ward horizontal  thrust.  Now  if  we  as- 
sume that  this  horizontal  thrust  is  at 
every  point  equal  in  intensity  to  the  ver- 
tical loading  at  that  point,  the  curve  of 
equilibrium  under  such  a  system  of 
forces  is  the  hydrostatic  curve.  This  is 
the  assumed  condition  of  the  forces  act- 
ing in  the  Neuilly  and  other  bridges  of 
this  class. 

When  the  spandrels  cannot  be  made 
firm  and  solid  this  form  should  not  be 
used,  but  when  they  can  be,  as  in  the 
successive  arches  of  a  stone  bridge,  it  is 
advantageous  rather  than  otherwise,  to 
have  such  a  thrust  from  the  arch  against 
the  spandrel ;  while  the  hydrostatic 


101 

curve  of  given  span  and  rise  gives  a 
greater  water-way  than  the  correspond- 
ing catenary  would. 

The  catenary  needs  no  resistance  from 
the  spandrel,  being  balanced  under  the 
vertical  load  alone. 

Example.  —  Let  the  span  be  100  ft. 
and  rise  30  ft.  Then  the  depth  of  load- 
ing at  the  crown  (=  A  Y,  Fig.  46)  will 
be  found  from  equation  52 


Then  OQ  =  91.7ft. 

Hence  H    =  wy^Q   (putting  w    =    160 
Ibs.)  =  107,600  Ibs. 
Depth  of  keystone 


=  A/TT2^T9r7  =  3.3  ft. 

This  gives  a  pressure  of  32,280  Ibs.  to 
the  sq.  ft.,  or  about  225  Ibs.  to  the  sq.  in. 
6.  If  the  vertical  forces  vary  as  in  the 
hydrostatic  arch,  and  the  horizontal  are 
not  equal  to  them,  but  differ  at  each 
point  in  a  constant  ratio,  the  curve  of 
equilibrium  (Fig.  47)  becomes  the  Geo- 


102 

static  curve  discussed  in  Case  VII.  This 
curve  derives  its  name  from  the  fact  that 
the  system  of  pressures  above  described 
is  similar  to  that  exerted  by  a  mass  of 
loose  earth  against  CAB.  Let  M  T  = 
the  horizontal  surface  of  the  earth  ;  then. 
FIG.  47. 

M  T 


at  each  point  D  of  the  arch  there  is  a 
vertical  pressure  of  intensity  (p'y)  pro- 
portional to  the  depth  (y)  of  D  below  M 
Y,  and  a  horizontal  pressure  whose  in- 
tensity is  less  than  p ' y  in  a  constant 
ratio, 
or  p'x  =  c*p'y 


103 

(c2  being  taken  to  represent   the  ratio  of 
the  intensities). 

Assume  a  hydrostatic  arch  whose  verti- 
cal dimensions  shall  be  identical  with 
those  of  the  geostatic  arch,  and  whose 
span  (C  B)  (Fig.  48)  shall  be  connected 


oo 


with  the  span  of  the  geostatic  arch  (C'  B') 
by  the  equation 


C'B' 


(55.) 


104 

The  intensity  of  the  vertical  pressure 
(the  horizontal  is  like  it)  in  this  hydro- 
static arch  must  be 


Py  = 

From  these  data  deduce  a  hydrostatic 
arch  and  then  pass  by  parallel  projec- 
tions to  the  required  geostatic  arch. 

Equations  (35)  (36)  (37)  (38)  give  the 
values  of  the  quantities  needed  in  dis- 
cussing the  Geostatic  arch. 

Example  1.  —  Let  the  span  of  the  geo- 
static arch  (C'  B'=  100  ft.)  be  given; 
also  the  depth  of  the  loading  (A  Y  =  20 
ft.)  ;  also  the  ratio  of  the  pressures  (c2  — 
J);  and  the  weight  of  a  cubic  ft.  of  the 
loading  =  w  —  100  Ibs.  Whence 

P'y*  =  Wo  =  2,000  Ibs. 
Then  since 

™  = 


c 
=  58  Ibs. 


"  .2000  =  l,154.71bs. 

We.  find  from  equations  (52)    (53)    (54) 
for  the  hydrostatic  arch 


105 

Else  =  a  =  O  A  =  57.7  ft. 

P0  =  140.93  ft. 

P!   =363  ft. 

H  =  V  =  T  =  pyo  Po  =  1,154.7  X  140.93  = 
162,700  Ibs.  nearly. 

In  the  geostatic  arch  we  have  from 
equations  (35)  (36)  (37)  and  (38) 

Thrust  at  B  =  V  =  V  =  162,700  Ibs. 

A  =  H'  =  cH  =  94,300  Ibs.  nearly. 
Po'  =46.97  ft.  p,'  =  62.65ft. 

Example  2.  —  Suppose  the  span  =  100 
ft.  depth,  A  Y  =  yQ  =  20  ft.  and  rise,  a 
=  30  ft.  given  ;  to  find  c  and  thence 
the  hydrostatic  arch. 

From  equation  (52)  we  find 

b  =  40.71, 

and   thence  in  same  equations  xl  =  39. 
Hence  the  span  of  the   hydrostatic  arch 

=  2xl  =  78  ft. 
Andasc.CB=  C'B' 


Then  proceed  as  in  the  last  example.  In 


106 

this  example  the  hydrostatic  arch  is  the 
smaller  of  the  two. 

The  line  of  pressures  in  a  geostatic 
arch  is  found  as  it  was  in  the  elliptic. 

The  geostatic  is  the  true  curve  of 
equilibrium  under  earth  pressure,  but 
when  A  Y  (Fig.  48)  is  great  compared 
with  A  O,  it  approximates  the  ellipse 
described  through  the  points  C'  A  B'  as 
already  stated. 

7.  Convenience,  or  other  reasons,  will 
often  dictate  the  form  of  the  arch  with- 
out reference  to  the  loading,  and  again 
necessity  may  make  the  vertical  load 
different  from  any  and  all  the  cases  we 
have  discussed.  In  such  instances  Case 
VIII.  will  enable  us  to  determine  the 
character  and  amount  of  the  horizontal 
forces  which  must  be  applied  through 
the  resistance  of  the  spandrel,  when  once 
the  form  of  the  arch  and  the  vertical 
load  are  known. 

When  the  horizontal  forces  thus  re- 
quired are  thrusts  directed  against  the 
arch,  it  is  generally  possible  so  to  build 
the  spandrel  that  the  arch  may  be  se- 


107 

cure,  but  when  they  are  the  opposite,  or 
outward  pulls  on  the  arch,  then  it  is 
difficult  to  insure  stability,  as  to  do  so 
requires  tension  between  the  arch  and 
the  spandrel.  In  such  cases  it  is  best 
to  change  the  form  of  the  arch. 


The  discussion  of  Case  VIII.  of  cords 
enables  us  to  determine  the  necessary 
data  in  the  case  of  similar  linear  arches 
under  similar  loads. 

Fig.  (50)   gives  the   geometrical  con- 


108 

struction  of  the  triangle  of  forces  at  every 
point  of  the  semi-arch  A  B  (Fig.  49). 

We  may  discuss  a  given  linear  arch 
CAB  under  a  given  vertical  load,  by  de- 
termining : 

1.  The  thrust  at  crown  ;  which  is 

H0=jpo/>0.  (560 

2.  Total  horizontal  thrust  required  on 
any  arc   A  D',    A  D",    etc.     This,  from 
equation  (42),  is 

H  =  H0  -  V  cot.  t.  (57.) 

If    this  be  negative  the  spandrel  must 
exert  a  pull  instead  of  a  thrust. 

On  the  half-arch  A  B  the  above  equa 
tion  becomes 

H!  =  H0  -  VT  cot.  »\.  (58.) 

On  any  arc  B  D'v,  counting  from  B 
upwards,  the  total  spandrel  thrust  is 

H,  -  H  =  —  Vi  cot.  a,  +  V  cot.  i.    (59.) 

This  last  expression  has  at  least  one 
maximum  value  corresponding  to  some 
arc  B  D.  In  the  Fig.  (49)  this  value 
corresponds  to  the  arc  B  D'". 


109 


I 


110 

Let  this  maximum  value  be  denoted  by 
Hm  and  let  im  =  the  inclination  at  D'". 
Then 

Hw  =  —  VA  cot.  t\  -f-  V  cot.  ew  =  E'"  G'". 
(Fig.  50)    (60.) 

~D"'  is  known  as  the  "  point  of  rupture." 
There  the  action  of  the  spandrel  ceases 
to  be  a  thrust,  and  must,  above  that 
point,  for  some  distance  at  least,  become 
tension. 

3.  The  intensity  of  the  horizontal 
spandrel  thrust  or  pull  in  any  layer  (as 
between  D'"  and  D'v)  is  from  equation 
(43) 

' 


d  cv  cot. 


_ 

dy   ~  dy  dy 

When  H  is  positive  (that  is  thrust)  px  is 
negative,  as  it  should  be,  since  it  is  equal 
to  the  increment  of  the  abscissas  of  the 
curve  F  G'  G",  etc.  (Fig.  50),  and  these 
increments  are  decreasing  from  G"'  to 
Gv. 

At  the  point  of  rupture 

#»  =  0.  (61.) 


Ill 

We  can  determine  the  point  of  rupture 
in  three  ways :  First,  by  constructing 
the  Fig.  (50)  and  finding  the  inclination 
(£m)  corresponding  to  the  maximum 
abscissa  E'"  G'".  Secondly,  by  substi- 
tuting the  various  values  of  i  and  V  in 
the  value  of 

(H  j  -  H)  (eq.  59), 

and  getting  the  maximum  value  of  the 
expression.  The  i  which  gives  this  max- 
imum value  corresponds  to  the  point  of 
rupture.  Thirdly,  by  solving  equation 
(61)  #,  =  0. 

4.  The  thrust  along   the  rib  at  every 
point  is  from  equation  40, 

T  =  V  cosec. »,  (62.) 

and  it  is  represented  by  the  inclined 
lines  F  S'  F  S",  etc.,  Fig.  (50). 

The     horizontal    component    of    this 
thrust  is 

Hr  =  V  cot.  i  —  the  abscissas  of  K  8',  etc. , 
which  are  always  equal  to  H0,  the  thrust 


112 

at  the  crown  minus  the  spandrel  thrust 
between  A  and  the  point  in  question 

.  •.  Hr  =  V  cot.  i  =  H0  -  H.     (63.) 

This  is  evidently  a  maximum  at  the  point 
of  rupture,  or,  since  at  the  point  of  rup- 
ture, 

H  =  HJ  —  Hm, 

we  have 

HB  =  H0  —  Hx  -|-  Hm. 

But 

H0  -  H!  =  Y!  COt.  i, 
...  HB  =  V,  cot.  i,  4-  Hm.         (64.) 

This  horizontal  thrust  of  the  rib  at  D'" 
is  therefore  to  be  balanced  by  the  hori- 
zontal reaction  of  the  abutment  at  B  (  — 
Vj  cot.  i^)  together  with  the  resistance 
of  the  spandrel  between  B  and  D'"  (  = 
Hm).  When  the  arch  is  vertical  at  B,  Vx 
cot.  iv=  0. 

5.  In  single  arches  it  is  necessary  to 
know  the  point  of  application  of  the  re- 
sultant of  the  forces  represented  by  (V, 
cot.  i^  -f  Hm)  in  order  to  determine  the 
stability  of  the  abutments.  Take  mo- 
ments with  reference  to  the  axis  of  ab- 


113 

scissas  M  Y.  Then  if  yB  =  ordinate  of 
point  in  question,  and  ym  and  yl  be  the 
ordinates  of  D'"  and  B,  we  have  * 


=  (Vt  cot.  i1)yl  +v&  H 

=  (Vl  cot.  *,)  yj  +  /    y  #B  dy. 

«/      ym 

/k 
ypxdy 
Vm 


HB 


(65.) 


In  this  we  neglect  the  spandrel  forces 
above  D'"  so  far  as  they  affect  the  sta- 
bility of  the  abutment.  This  can  be 
done  with  safety. 

The  line  of  pressures  and  depth  of 
keystone  are  determined  as  heretofore. 

Example  1. — Let  the  assumed  form  of 
the  soffit  be  a  semi-circle,  and  let  the 
loading  consist  of  the  arch  and  backing 
of  homogeneous  masonry  carried  up  to  a 
horizontal  "extrados"  M  Y  (Fig.  51). 

Place  the  radius  of  the  arch  —  r 

Depth  A  Y  =  a  r 
Heaviness  of   the  material   —  w 


114 

Take  the  origin  of  co-ordinates  at  A  and 


1C 

6 


* 

express  the  co-ordinates  in  terms  of  the 


115 

inclination  i  of   the   arch  as    on  p.  217 

RanJcinds  C.  E. 

Then 

Thrust  at  crown  =  H0  —  p0  p0  =  (war)  v  = 
war 

Vertical  load  on  any  arc  =  V  =  wr2 

cos.  *  sin.  i       i 


Spandrel  thrust  on  any  arc  A  D 
H  =  H0  -  V  cot.  i  =  wr2 

(  .  cos.  2  i  .  i  cos.  i  ) 

1  a  -  (1  +  a)  cos.  i  -\  --  s  --  {-  =—.  —  .  I 
]  2         '  2  sin.  i  j 

On  A  B  this  becomes  (since  the  arch  is 
vertical  at  B  and  C) 

H!  =  war9  =  H0 
.  '  .  Hm  —  V  cot.  im 

Intensity  of  spandrel  thrust 
d  (V  cot.  0 

-   --" 


,.  t  —  cos.  i  sin.  t 

(1   +  a)  -  cos.  .  - 


The  point   of  rupture  is   found  by  put- 


116 

ting  px  =  o  and  finding  the  value  of   im 
by  trials.      As  a  first  approximation 

1  +  30 
im  =  arc.  cos.  - 

a 

Thrust  along  the  rib  =  T  =  V  cosec  i. 
At  B  this  is 

Y!  —  tw»/a+l  -      -V 

So 

HB  —  Y!  cot.  e\  +  B.m  =  Hm  =  wr8 


cos.  im  - 


and 

ra     /»90° 

^R  —  HR"  /    •      PX   sin'  *  (!  —  cos-  0 
47    ^m 

^Example  2.  —  Let 

r  =  20'  A  Y  =  2.5'. 

Then 

25       1 

*  =  "io  =  8  w  =  15°  lbs-    (Fig- 

Then 

H0  =  ^«^=  7,500  lbs. 

(  9    .  cos.  i  sin.  £        ?, 

V  -  60,000        sin.*-       -  ---- 


117 


At  B,    V  =  60,000        -  --  ~     =  20,376 Ibs. 


s 


Angle  of  rupture 
im  —  arc.  cos.  -^- 


.-1. 6875  =  46°  34' 


118 


HR  =  Hm  =  60,000 


_ 

8 


.Six. 947  |  =  8>154 
2         j"      Ibs. 


(Fig.  53)  shows  the   manner  in  which 


the  forces  vary.     From  A  to  D  (Fig.  52) 
there  must  be  a  pull  in   the  spandrel  to 


119 

produce  equilibrium.     The  total  amount 
of  this  pull  is  small,  being 

=8,154  -  7,500  =  654  Ibs. 

To  rid  the  arch  of  it,  so  that  the  part 
D1  A  D  shall  either  be  balanced  under 
the  vertical  load  alone  or  exert  a  thrust 
outwards,  instead  of  a  pull  inwards,  we 
flatten  the  arc  D1  A  D.  A  few  trials 
will  determine  this  flattening  near  enough 
for  practice. 

Thus  if  D1  A  D  (Fig.  54)  is  to  be  bal- 
anced under  the  vertical  load  alone,  find 
the  center  of  gravity  of  the  section  D1  A 
and  its  load.  Draw  a  vertical  line  P 
through  this  point,  then  if  we  can  draw 
a  line  from  any  point  in  the  middle 
third  of  the  joint  D1  parallel  to  the  tan- 
gent to  the  arch  there,  and  from  its  in- 
tersection with  P  draw  a  line  parallel  to 
the  arch  at  A  which  will  intersect  A  L 
within  the  middle  third,  then  the  extreme 
points  of  the  line  of  pressures  in  the 
section  A  D1  will  be  within  the  middle 
third,  and  the  line  of  pressures  will  gen- 
erally be  altogether  within  it. 


FIG.  54. 


121 

The  new  radius  required  for  the  arc 
D1  A  D  may  also  be  determined  roughly 
by  putting  H0  =  war*  =  HR  and  thence 
getting  r1,  since,  if  D1  A  D  is  to  be  bal- 
anced under  vertical  load  alone,  the  hori- 
zontal thrust  at  every  point  of  it  must  be 
the  same  and  =  HB,  the  thrust  at  D1 
andD. 


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SCIENTIFIC    PUBLICATION.  35 

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Amply   Illustrated  when  the  Subject   Demands. 

No.  1.— CHIMNEYS  FOB  FURNACES  AND  STEAM 
BOILERS.  By  R.  Armstrong,  C.  E.  Third 
American  edition.  Revised  and  partly  rewritten, 
with  an  Appendix  on  Theory  of  Chimney  Draught, 
by  F.  E.  Idell,  M.  E. 

No.  2.— STEAM  BOILEB  EXPLOSIONS.  By  Zerah 
Colburn.  New  edition,  revised  by  Prof.  R.  H. 
Thurston. 

No.  3.— PRACTICAL  DESIGNING  OF  BETAINING- 
WALLS.  By  Arthur  Jacob,  A.  B.  Second  edition, 
revised,  with  additions  by  Prof.  W.  Cain. 

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BBIDGES.  By  Chas.  E.  Bender,  C.  E.  Second 
edition,  with  appendix. 

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Butler.  Second  edition,  re-edited  and  enlarged  by 
James  L.  Greenleaf ,  C.  E. 

Xo.  6.— ON  THE  DESIGNING  AND  CONSTBUCTION 
OF  STORAGE  BESERVOIRS.  By  Arthur 
Jacob,  A.  B.  Second  edition,  revised,  with  addi- 
tions by  E.  Sherman  Gould. 

No.  7.— SURCHARGED  AND  DIFFEBENT  FORMS 
OF  BETAINING-WALLS.  By  James  S.  Tate, 
C.  E. 

No.  8.— A  TBEATISE  ON  THE  COMPOUND  EN- 
GINE. By  John  Turnbull,  Jun.  Second  edition, 
revised  by  Prof.  S.  W.  Robinson. 

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C.  E.  Founded  on  tbe  original  treatise  of  C, 
William  Siemens,  D.  C.  L. 

No.  10.— COMPOUND  ENGINES.  Translated  from  th» 
French  of  A.  Mallet.  Second  edition,  revised  with  re- 
sults of  American  Practice,  by  Richard  H.  Buel,!C.  E. 


2  THE   VAN   NOSTRAND   SCIENCE    SERIES. 

No.  11.— THEORY   OF    ARCHES.    By  Prof.  W.  Allan. 

No.  12.— THEORY  OF  VOUSSOIR  ARCHES.  By  Prof. 
W.  G.  Cain.  Second  edition,  revised  and  enlarged. 

No.  13.-GASES  MET  WITH  IN  COAL  MINES.  By 
J.  J.  Atkinson.  Third  edition,  revised  and  enlarged 
by  Edward  H.  Williams,  Jun. 

No.  14.— FRICTION  OF  AIR  IN  MINES.  By  J.  J. 
Atkinson.  Second  American  edition. 

No.  15.— SKEW  ARCHES.  By  Prof.  E.  W.  Hyde,  C.  E. 
Illustrated.  Second  edition. 

No.  16.— A  GRAPHIC  METHOD  FOR  SOLVING 
CERTAIN  ALGEBRAIC  EQUATIONS.  By 
Prof.  G.  L.  Vose. 

No.  17.— WATER  AND  WATER-SUPPLY.  By  Prof. 
W.  H.  Corfield,  of  the  University  College,  London. 
Second  American  edition. 

No.  18.— SEWERAGE  AND  SEWAGE  PURIFICA- 
TION. By  M.  N.  Baker,  Associate  Editor  "En- 
gineering News." 

No.  19.— STRENGTH  OF  BEAMS  UNDER  TRANS- 
VERSE LOADS.  By  Prof.  W.  Allan,  author  of 
"  Theory  of  Arches."  Second  edition,  revised. 

No.  20.— BRIDGE  AND  TUNNEL  CENTRES.  By 
John  B.  McMaster,  C.  E.  Second  edition. 

No.  21.— SAFETY  VALVES.  Third  edition.  By  Richard 
H.  Buel,  C.  E. 

No.  22. -HIGH  MASONRY  DAMS.  By  E.  Sherman 
Gould,  M.  Am.  Soc.  C.  E. 

No.  23.— THE  FATIGUE  OF  METALS  UNDER  RE- 
PEATED STRAINS.  With  various  Tables  of 
Results  and  Experiments.  From  the  German  of 
Prof.  Ludwig  Spangenburgh,  with  a  Preface  by  S. 
H.  Shreve,  A.  M? 

No.  24.— A  PRACTICAL  TREATISE  ON  THE 
TEETH  OF  WHEELS.  By  Prof.  S.  W.  Robin- 
son. Second  edition,  revised. 

No.  25.— ON  THE  THEORY  AND  CALCULATION 
OF  CONTINUOUS  BRIDGES.  By  Mansfield 
Merriman,  Ph.  D. 

No.  26.— PRACTICAL  TREATISE  ON  THE  PROP- 
ERTIES OF  CONTINUOUS  BRIDGES.  By 
Charles  Bender,  C.  E. 


THE   VAN   NOSTRAND    SCIENCE   SERIES.  3 

No.  27.— ON  BOILER  INCRUSTATION  AND  COR- 
ROSION. By  F.  J.  Rowan.  New  edition.  Revised 
by  F.  E.  Idell. 

No.  28.— TRANSMISSION  OF  POWER  BY  WIRE 
ROPES.  Second  edition.  By  Albert  W.  Stahl, 
U.  S.  N. 

No.  29.— STEAM  INJECTORS.  Translated  from  the 
French  of  M.  Leon  Pochet. 

No.  30.— TERRESTRIAL  MAGNETISM  AND  THE 
MAGNETISM  OF  IRON  VESSELS.  By  Prof. 
Fairman  Rogers. 

No.  31.— THE  SANITARY  CONDITION  OF  DWELL- 
ING-HOUSES IN  TOWN  AND  COUNTRY. 
Second  edition,  revised.  By  George  E.  Waring,  Jun. 

No.  32.— CABLE  -  MAKING  FOR  SUSPENSION 
BRIDGES.  By  W.  Hildebrand,  C  E. 

No.  33.— MECHANICS  OF  VENTILATION.  By 
George  W.  Rafter,  C.  E.  New  and  Revised  Edition. 

No.  34.— FOUNDATIONS.  By  Prof.  Jules  Gaudard,  C.  E. 
Second  edition.  Translated  from  the  French. 

No.  35.— THE  ANEROID  BAROMETER  :  ITS  CON- 
STRUCTION AND  USE.  Compiled  by  George 
W.  Plympton.  Seventh  edition,  revised  and 
enlarged. 

No.  36. —MATTER  AND  MOTION.  By  J.  Clerk  Max- 
well, M.  A.  Second  American  edition. 

No.  37.— GEOGRAPHICAL  SURVEYING;  ITS  USES,. 
METHODS  AND  RESULTS.  By  Frank  D& 
Yeaux  Carpenter,  C.  E. 

No.  38.— MAXIMUM  STRESSES  IN  FRAMED 
BRIDGES  By  Prof.  William  Cain,  A.  M.,  C.  E. 
New  and  revised  edition. 

No.  39.— A  HAND-BOOK  OF  THE  ELECTRO- 
MAGNETIC TELEGRAPH.  By  A.  E.  Loring. 

No.  40.— TRANSMISSION  OF  POWER  BY  COM- 
PRESSED AIR.  By  Robert  Zahner,  M.  E. 
Second  edition. 

No.  41.— STRENGTH  OF  MATERIALS.  By  William 
Kent,  C.  E.,  Assoc.  Editor,  "  Engineering  News.'* 
Second  edition. 


4  THE   VAN   NOSTRAND    SCIENCE   SERIES. 

No.  42.— VOUSSOIR  ARCHES  APPLIED   TO  STONE 
BRIDGES,     TUNNELS,     CULVERTS,     AND 

DOMES.    By  Prof.  William  Cain. 
No.  43  —  WAVE    AND    VORTEX    MOTION.      By    Dr. 

Thomas  Craig,  of  Johns  Hopkins  University. 
No.  44— TURBINE    WHEELS.     By  Prof.   W.  P.  Trow- 

bridge,  Columbia  College.  Second  edition.  Revised. 
No.  45.— THERMO— DYNAMICS.     By  Prof.  H.  T.  Eddy, 

University  of  Cincinnati. 
No   46.— ICE-MAKING  MACHINES     From  the  French 

of  M.  Le  Doux.    Revised  by  Prof.  Denton.    Fourth 

edition. 
No.  47.— LINKAGES  ;    THE      DIFFERENT    FORMS 

AND     USES     OF     ARTICULATED    LINKS. 

By  J.  D.  C.  de  Roos. 
No.  48.— THEORY       OF      SOLID      AND      BRACED 

ARCHES.    By  William  Cain,  C.  E. 
No.  49.— ON    THE    MOTION    OF    A    SOLID    IN    A 

FLUID.    By  Thomas  Craig,  Ph.  D. 
^0.  50.— DWELLING-HOUSES  :    THEIR  SANITARY 

CONSTRUCTION  AND   ARRANGEMENTS. 

By  Prof.  W.  H.  Corfield. 
No.  51.— THE    TELESCOPE  :    ITS    CONSTRUCTION, 

Etc.    By  Thomas  Nolan. 
No.  52.— IMAGINARY  QUANTITIES.  Translated  from 

the  French  tf  M.  Argand.    By  Prof.  Hardy. 
No.  53.— INDUCTION     COILS  :    HOW    MADE    AND 

HOW  USED.    Ninth  Edition. 
No.  54.— KINEMATICS    OF   MACHINERY.    By  Prof. 

Kennedy.    With  an  introduction  by  Prof.  R.   H. 

Thurston. 
No.  55.-SEWER   GASES :     THEIR    NATURE    AND 

ORIGIN.     By  A.    de  Varona.     Second   edition, 

revised  and  enlarged. 
y0t  56.— THE    ACTUAL    LATERAL    PRESSURE  OF 

EARTHWORK.     By  Benj.  Baker,  M.  Inst.,  C.  E. 
-$0t  57.— INCANDESCENT    ELECTRIC    LIGHTING. 

A  Practical  Description  of  the  Edison  System.    By 

L.  H.  Latimer,  to  which  is  added  the  design  and 

Operation  of  Incandescent  Stations.  By  C.  J.  Field, 

and   the   Maximum     Efficiency   of   Incandescent 

Lamps,  by  John  W.  Howell. 


THE   VAN   NOSTRAND    SCIENCE    SERIES.  5 

No.  58.— THE     VENTILATION    OF    COAL,     MINES. 

By  W.  Fairley,  M.  E.,  F.  S.  S.,  and  Geo.  J.  Andre. 
No.  59.— RAILROAD      ECONOMICS  ;      OR,      NOTES 

WITH  COMMENTS.     By  S.W.  Robinson,  C.  E. 
Xo.  60.— STRENGTH  OF  WROUGHT-IRON  BRIDGE 

MEMBERS.    By  S.  W.  Robinson,  C.  E. 
No.  61.— POTABLE    WATER    AND    METHODS     OF 

DETECTING  OIPURITIES.   By  M.  N.  Baker. 
No.  62.— THE  THEORY  OF  THE  GAS-ENGINE.    By 

Dougald  Clerk.    Second  edition.    With  additional 

matter.    Edited  by  F.  E.  Idell,  M.  E. 
No.  63.— HOUSE      DRAINAGE      AND       SANITARY 

PLUMBING.    By  W.  P.  Gerhard.  Eighth  edition. 

Revised. 
No.  64.— ELECTRO-MAGNETS.       By    Th.   du    Moncel. 

Second  revised  edition. 
No.  65.— POCKET  LOGARITHMS  TO  FOUR  PLACES 

OF  DECIMALS. 
No.  66.— DYNAMO-ELECTRIC  MACHINERY.  ByS.P. 

Thompson.  With  notes  by  F.  L.  Pope .  Third  edition. 
No.  67.— HYDRAULIC  TABLES  BASED  ON   "  KUT- 

TER'S  FORMULA."    By  P.  J.  Flynn. 
No.  68.— STEAM-HEATING.    By  Robert  Briggs     Third 

edition,  revised,  with  additions  by  A.  R.  Wolff. 
No.  69.— CHEMICAL  PROBLEMS.  By  Prof.  J.  C.  Foye. 

Fourth  edition,  revised  and  enlarged. 

No.  70— EXPLOSIVE  MATERIALS.    By  M.  Bertholet . 
No.  71.— DYNAMIC  ELECTRICITY.    By  John  Hopkin- 

son,  J.  A.  Schoolbred,  and  R.  E.  Day. 
lSTo.  72.— TOPOGRAPHIC AL  SURVEYING.    By  George 

J.  Specht,  Prof.  A.  S.  Hardy,  John  B.  McMaster, 

and  H.  F.  Walling.    Second  edition,  revised. 
No.  73.— SYMBOLIC    ALGEBRA;  OR,    THE    ALGE- 
BRA OF  ALGEBRAIC  NUMBERS.    By  Prof. 
W.  Cain. 

No.  74.— TESTING  MACHINES:  THEIR  HISTORY, 
CONSTRUCTION,  AND  USE.  By  Arthur  V. 
Abbott. 

3Jo.  75.— RECENT  PROGRESS  IN  DYNAMO-ELEC- 
TRIC MACHINES.  Being  a  Supplement  to  Dy- 
namo-Electric Machinery.  By.  Prof.  Sylvanus  P. 
Thompson. 


6  THE   VAN   NOSTRAND    SCIENCE   SERIES. 

No.  76.— MODERN  REPRODUCTIVE  GRAPHIC 
PROCESSES.  By  Lieut.  James  S.  Pettit,  U.S.A. 

No.  77.— STADIA  SURVEYING.  The  Theory  of  Stadia 
Measurements.  By  Arthur  Winslow. 

No.  78.— THE  STEAM-ENGINE  INDICATOR,  AND 
ITS  USE.  ByW.  B.LeVan. 

No.  79.— THE  FIGURE  OF  THE  EARTH.  By  Frank 
C.  Roberts,  C.  E. 

No.  80.— HEALTHY  FOUNDATIONS  FOR  HOUSES. 
By  Glenn  Brown. 

No.  81.— WATER  METERS  :  COMPARATIVE  TESTS 
OF  ACCURACY,  DELIVERY,  Etc.,  Distinc- 
tive features  of  the  Worthington,  Kennedy,  Sie- 
mens, and  Hesse  meters.  By  Ross  E.  Browne. 

No.  82.— THE  PRESERVATION  OF  TIMBER  BY 
THE  USE  OF  ANTISEPTICS.  By  Samuel 
Bagster  Boulton,  C.  E. 

No.  83.— MECHANICAL  INTEGRATORS.  By  Prof. 
Henry  S.  H.  Shaw,  C.  E. 

No.  84.— FLOW  OF  WATER  IN  OPEN  CHANNELS, 
PIPES,  CONDUITS,  SEWERS,  Etc.  With 
Tables.  By  P.  J.  Flynn,  C.  E. 

No.  85.— THE  LUMINIFEROUS  AETHER.  By  Prof, 
de  Volson  Wood. 

No.  86.— HAND-BOOK  OF  MINERALOGY;  DETER- 
MINATION AND  DESCRIPTION  OF  MIN- 
ERALS FOUND  IN  THE  UNITED  STATES. 
By  Prof.  J.  C.  Foye,  Sixth  edition,  revised. 

No.  87.— TREATISE  ON  THE  THEORY  OF  THE 
CONSTRUCTION  OF  HELICOIDAL  OB- 
LIQUE ARCHES.  By  John  L.  Culley,  C.  E. 

No.  88.— BE AMS  AND  GIRDERS.  Practical  Formulas 
for  their  Resistance.  By  P.  H.  Philbrick. 

No.  89.— MODERN  GUN  COTTON :  ITS  MANUFAC- 
TURE, PROPERTIES,  AND  ANALYSIS.  By 
Lieut.  John  P.  Wisser,  U.S.A. 

No.  90.— ROTARY  MOTION  AS  APPLIED  TO  THE 
GYROSCOPE.  By  Gen.  J.  G.  Barnard. 

No.  91.-LEVELING:  BAROMETRIC  TRIGONOME- 
TRIC, AND  SPIRIT.  By  Prof.  I.  O.  Baker. 

No.  92.— PETROLEUM :  ITS  PRODUCTION  AND 
USE.  By  Boverton  Redwood,  F.  I.  C.,  F.  C.  S. 


THE  VAN   NOSTRAND   SCIENCE   SERIES.  7 

No.  93. -RECENT  PRACTICE  IN  THE  SANITARY 
DRAINAGE  OF  BUILDINGS.  With  Memor- 
anda on  the  Cost  of  Plumbing  Work.  Second  edi- 
tion, revised.  By  William  Paul  Gerhard,  C.  E. 

No.  94.— THE  TREATMENT  OF  SEWAGE.  By  Dr.  C. 
Meymott  Tidy. 

No.  95.— PLATE  GIRDER  CONSTRUCTION.  By  Is- 
ami  Hiroi,  C.  E.  2d  edition,  revised  and  enlarged. 

No.  96.— ALTERNATE  CURRENT  MACHINERY.  By 
Gisbert  Kapp,  Assoc.  M.  Inst.,  C.  E. 

No.  97.— THE  DISPOSAL  OF  HOUSEHOLD  WASTES. 
By  W.  Paul  Gerhard,  Sanitary  Engineer. 

No.  98.— PRACTICAL  DYNAMO  BUILDING  FOR 
AMATEURS.  HOW  TO  WIND  FOR  ANY 
OUTPUT.  By  Frederick  Walker.  Fully  illus- 
trated. 

No,  Q9.— TRIPLE-EXPANSION  ENGINES  AND  EN- 
GINE TRIALS.  By  Prof.  Osborne  Reynolds. 
Edited  with  notes,  etc.,  by  F.  E.  Idell,  M.  E. 

No.  100.— HOW  TO  BECOME  AN  ENGINEER,  or  the 

Theoretical  and  Practical  Training  necessary  in 
fitting  for  the  duties  of  the  Civil  Engineer.  By 
Prof.  Geo.  W.  Plympton. 

No.  101.— THE  SEXTANT,  and  other  Reflecting  Mathe- 
matical Instruments.  With  Practical  Hints  for 
their  adjustment  and  use.  By  F.  R.  Brainard,  U. 
S.  Navy. 

No.  102.— THE  GALVANIC  CIRCUIT  INVESTI- 
GATED MATHEMATICALLY.  By  Dr.  G.  S. 
Ohm,  Berlin,  1827.  Translated  by  William  Fran- 
cis.  With  Preface  and  Notes  by  the  Editor, 
Thomas  D.  Lockwood,  M.  I.  E.  E. 

No.  103.— THE    MICROSCOPICAL    EXAMINATION 

OF  POTABLE  WATER.  With  Diagrams.  By 
Geo.  W.  Rafter. 

1*0.  104.— VAN  NOSTRAND'S  TABLE  BOOK  FOR 
CIVIL  AND  MECHANICAL  ENGINEERS. 

Compiled  by  Prof.  Geo.  W.  Plympton. 

No.  105.— DETERMINANTS.  An  Introduction  to  the 
Study  of,  with  Examples  and  Applications.  By 
Prof.  G.  A.  Miller. 


8  THE  VAN   NOSTRAND    SCIENCE   SERIES. 

No.  106.— COMPRESSED  AIR.  Experiments  upon  the 
Transmission  of  Power  by  Compressed  Air  in 
Paris.  (Popp's  System.)  By  Prof  A.  B.  W.  Ken 
nedy.  The  Transmission  and  Distribution  of 
Power  from  Central  Stations  by  Compressed  Air. 
By  Prof.  W.  C.  Unwin. 

No.  107.— A  GRAPHICAL  METHOD  FOR  SWING 
BRIDGES.  A  Rational  and  Easy  Graphical  An- 
alysis of  the  Stresses  in  Ordinary  Swing  Bridges. 
Witn  an  Introduction  on  the  General  Theory  of 
Graphical  Statics.  By  Benjamin  F.  La  Rue.  4 
Plates. 

No.  108.-SLIDE  VALVE  DIAGRAMS.  A  French  Method 
for  Constructing  Slide  Valve  Diagrams.  By  Lloyd 
Bankson,  B.  S.,  Assistant  Naval  Constructor,  U.  S. 
Navy.  8  Folding  Plates. 

No.  109.— THE  MEASUREMENT  OF  ELECTRIC 
CURRENTS.  Electrical  Measuring  Instruments. 
By  James  Swinburne.  Meters  for  Electrical  En- 
ergy. By  C.  H.  Wordingham.  Edited,  with  Pre- 
face, by  T.  Commerford  Martin.  Folding  Plate 
and  numerous  illustrations. 

No.  110. -TRANSITION  CURVES.  A  Field-Book  for 
Engineers,  containing  Rules  and  Tables  for  Laying 
out  Transition  Curves.  By  Walter  G.  Fox,  C.  E. 

No.  111.— GAS  -  LIGHTING  AND  GAS  -  FITTING. 
Specifications  and  Rules  for  Gas-Piping.  Notes  on 
the  advantages  of  Gas  for  Cooking  and  Heating, 
and  Useful  Hints  to  Gas  Consumers.  Second  edi- 
tion, rewritten  and  enlarged.  By  Wm.  Paul  Ger- 
hard, C.  E. 

No.  112.— A   PRIMER    ON   THE    CALCULUS.     By  E. 
Sherman  Gould,  M.  Am.  Soc.  C.  E.    Second  edi-  < 
tion,  revised  and  enlarged. 

No.  113.— PHYSICAL  PROBLEMS  and  their  Solution. 
By  A.  tourgougnon,  Formerly  Assistant  at  Belle- 
vue  Hospital. 

No.  114.— MANUAL  OF  THE  SLIDE  RULE.  By  F.  A. 
Halsey,  of  the  American  Machinist. 

No.  115.— TRAVERSE  TABLE  showing  the  difference- 
of  Latitude  and  Departure  for  distances  between 
1  and  100  and  for  Angles  to  Quarter  Degrees  be- 
tween 1  degree  and  90  degrees.  (Reprinted  from 
Scribner  s  Pocket  Table  Book  ) 


U.C.  BERKELEY 


No.  10 


No.  10} 


Navy. 


IN  BRIDGE  MEM- 
E. 

ODS  OF  DETECT- 
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GAS  -  ENGINE.  By 
nn.<iid  edition.  With  additional 
/F.  E.  Idell,  M.E. 


CODbOf 


2.—  THE     GAJXGE  AND    SANITARY    PLUMB- 
p   \V.  P.  Gerhard.      Sixth  edition.      Re- 


No. 


No.  64.— ELECTRO-MAGNETO.  3y  Th.  du  Moncel.  2d  re- 
vised edition. 

No.  65.— POCKET  LOGARITHMS  TO  FOUR  PLACES  OP 
DECIMALS. 

No.  66.— DYNAMO-ELECTRIC   MACHINERY.     By    S.    P. 

Thompson.    With  notes  by  F.  L.   Pope.    Third 

edition. 
No.  67.-HYDRAULIC  TABLES  BASED  ON    "KUTTER'S 

FORMULA."    By  P.  J.  Flynn. 
No.  68.— STEAM-HEATING.     By  Robert    Briggs.       Third 

edition,  revised,  with  additions  by  A.  R.  Wolff. 
No.  69.— CHEMICAL   PROBLEMS.    By  Prof.  J.  C.  Foye. 

Tkird  edition,  revised  and  enlarged. 
Xo    70.— EXPLOSIVE  MATERIALS.    By  M.  Bertholet. 

No.  71.— DYNAMIC  ELECTRICITY.    By  John  Hopkinson, 

J.  A.  Schoolbred,  and  R.  E.  Day. 
No.  72.— TOPOGRAPHICAL    SURVEYING.     By  George  J. 

Specht,  Prof.  A.  S.  Hardy,  John  B.  McMaster,  and 

H.  F.  Walling. 
No.  73.— SYMBOLIC  ALGEBRA:  OR,  THE  ALGEBRA  OF 

ALGEBRAIC  NUMBERS.     By  Prof.  W.  Cain. 
No.  74.-TESTING  MACHINES  :  THEIR  HISTORY,  CON- 

STRUCT1ON,  AND  USE.     By  Arthur  V.  Abbott. 
No.  75.— RECENT    PROGRESS    IN    DYNAMO-ELECTRIC 

MACHINES.     Beinsr  a  Supplement  to  Dynamo- 

Electric     Machinery.      By    Prof.     Sylvanus    P. 

Thompson. 

No.  76.— MODERN  REPRODUCTIVE  GRAPHIC  PRO- 
CESSES. By  Lieut.  James  S,  Pettit,  U.S.A. 

No.  77.— STADIA  SURVEYING.  The  Theory  of  Stadia 
Measurements.  By  Arthur  Winslow. 

No.  78.— THE  STEAM-ENGINE  INDICATOR,  AND  ITS 
USE.  By  W.  B.  Le  Van. 

No.  79.— THR  FIGURE  OF  THE  EARTH.  By  Frank  O. 
Roberts,  C.E. 

No.  80.-HEALTH1  FOUNDATIONS  FOR  HOUSES.  3y 
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A  Rational  and  Easy   Graphr    A 1 1 

Stresses  in   Ordinary   Swir 

Introduction  on  the  Gener^ 

Statics.    By  Benjamin  W 
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for  ConsLiuctio    Slid* 

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